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Learn to remove minimum elements so two arrays have no common elements in C: two-map and single-map O(n+m) approaches with code examples.
C Programming

Remove Minimum Elements so No Common Elements in C

Remove Minimum Elements so No Common Elements Exist in Two Arrays (C Tutorial)

Given two arrays, find the minimum number of elements that must be removed (across both arrays combined) so that the resulting arrays have no common elements. This problem is a subtle twist on intersection counting — and the solution is remarkably elegant.

This tutorial covers the insight, two C implementations (two maps vs. single map), edge cases, and best practices.

Conceptual Introduction & Fundamentals

What is the Problem?

Given two arrays arr1[] (size n) and arr2[] (size m), find the minimum number of elements to remove (from either array) so that no value appears in both arrays.

Examples:

text

Input: arr1 = [2, 3, 4, 5, 8], arr2 = [1, 2, 3, 4] Output: 3 (remove 2, 3, 4 from one array) Input: arr1 = [1, 2, 3, 4], arr2 = [5, 6, 7] Output: 0 (no common elements)

Why This Problem Matters

  • Set operations: Deepens understanding of intersection.
  • Multiset reasoning: Frequencies matter, not just presence.
  • Two implementation strategies: Two maps vs. one map.
  • Real-world use: Data deduplication, cache invalidation, conflict resolution.
  • Interview preparation: Common coding interview question.

Key Insight

For each value v present in both arrays with frequencies f1(v) in arr1 and f2(v) in arr2, we must remove all copies from one side (or split removals) so no copy remains in both arrays. The cheapest approach is to remove min(f1(v), f2(v)) copies — remove all copies from the side with fewer.

Summing over all common values:

text

answer = Σ over common values v of min(f1(v), f2(v))

Approaches Overview

ApproachTimeSpace
Two hash mapsO(n + m)O(n + m)
Single hash mapO(n + m)O(n)

Syntax & Core Mechanics

Approach 1: Two Hash Maps

Count frequencies in both arrays; sum min(f1, f2) for common values.

c

#include <stdio.h> #include <stdlib.h> #define TABLE_SIZE 4096 typedef struct { int *keys; int *counts; int *occupied; int capacity; } FreqMap; static unsigned int hashFunc(int key, int cap) { return ((unsigned int)key) % (unsigned int)cap; } FreqMap *createMap(int cap) { FreqMap *m = malloc(sizeof(FreqMap)); if (m == NULL) return NULL; m->keys = malloc((size_t)cap * sizeof(int)); m->counts = calloc((size_t)cap, sizeof(int)); m->occupied = calloc((size_t)cap, sizeof(int)); if (m->keys == NULL || m->counts == NULL || m->occupied == NULL) { free(m->keys); free(m->counts); free(m->occupied); free(m); return NULL; } m->capacity = cap; return m; } void freeMap(FreqMap *m) { if (m) { free(m->keys); free(m->counts); free(m->occupied); free(m); } } void increment(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { m->counts[idx]++; return; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; } m->keys[idx] = key; m->counts[idx] = 1; m->occupied[idx] = 1; } int getCount(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) return m->counts[idx]; idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) break; } return 0; } long long minRemoveTwoMaps(const int a[], int n, const int b[], int m) { FreqMap *ma = createMap(TABLE_SIZE); FreqMap *mb = createMap(TABLE_SIZE); if (ma == NULL || mb == NULL) { freeMap(ma); freeMap(mb); return -1; } for (int i = 0; i < n; i++) increment(ma, a[i]); for (int i = 0; i < m; i++) increment(mb, b[i]); long long res = 0; for (int i = 0; i < ma->capacity; i++) { if (!ma->occupied[i]) continue; int v = ma->keys[i]; int c2 = getCount(mb, v); if (c2 > 0) { int c1 = ma->counts[i]; res += (c1 < c2) ? c1 : c2; } } freeMap(ma); freeMap(mb); return res; } int main(void) { int arr1[] = {2, 3, 4, 5, 8}; int arr2[] = {1, 2, 3, 4}; printf("%lld\n", minRemoveTwoMaps(arr1, 5, arr2, 4)); return 0; }

Expected output:

text

3

Time: O(n + m), Space: O(n + m)

Approach 2: Single Hash Map (Optimal)

Count arr1 frequencies in one map; iterate arr2; when a value with positive count is found, decrement and add 1.

c

long long minRemoveSingleMap(const int a[], int n, const int b[], int m) { FreqMap *count = createMap(TABLE_SIZE); if (count == NULL) return -1; for (int i = 0; i < n; i++) increment(count, a[i]); long long res = 0; for (int i = 0; i < m; i++) { int c = getCount(count, b[i]); if (c > 0) { /* Decrement manually */ unsigned int idx = hashFunc(b[i], count->capacity); int start = (int)idx; while (count->occupied[idx]) { if (count->keys[idx] == b[i]) { count->counts[idx]--; break; } idx = (idx + 1) % (unsigned int)count->capacity; if ((int)idx == start) break; } res++; } } freeMap(count); return res; }

Time: O(n + m), Space: O(n)

Step-by-Step Trace for arr1 = [1,2,3,4], arr2 = [2,3,4,5,8]

text

Insert into count: {1:1, 2:1, 3:1, 4:1} Iterate arr2: 2 → count[2] = 1 > 0 → decrement to 0, res = 1 3 → count[3] = 1 > 0 → decrement to 0, res = 2 4 → count[4] = 1 > 0 → decrement to 0, res = 3 5 → count[5] = 0 → skip 8 → count[8] = 0 → skip Return 3 ✓

Full Practical Code Example with Explanation

This program implements both approaches, cross-verifies them, and exercises edge cases.

c

/** * min_remove_no_common.c * Minimum elements to remove so no common elements exist. * Compares two-map and single-map approaches. */ #include <stdio.h> #include <stdlib.h> #include <string.h> #include <errno.h> #include <time.h> #define MAX_LINE_LEN 100 #define TABLE_SIZE 262144 /* ============================================================ * FreqMap * ============================================================ */ typedef struct { int *keys; int *counts; int *occupied; int capacity; } FreqMap; static unsigned int hashFunc(int key, int cap) { return ((unsigned int)key) % (unsigned int)cap; } FreqMap *createMap(int cap) { FreqMap *m = malloc(sizeof(FreqMap)); if (m == NULL) return NULL; m->keys = malloc((size_t)cap * sizeof(int)); m->counts = calloc((size_t)cap, sizeof(int)); m->occupied = calloc((size_t)cap, sizeof(int)); if (m->keys == NULL || m->counts == NULL || m->occupied == NULL) { free(m->keys); free(m->counts); free(m->occupied); free(m); return NULL; } m->capacity = cap; return m; } void freeMap(FreqMap *m) { if (m) { free(m->keys); free(m->counts); free(m->occupied); free(m); } } void increment(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { m->counts[idx]++; return; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; } m->keys[idx] = key; m->counts[idx] = 1; m->occupied[idx] = 1; } int getCount(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) return m->counts[idx]; idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) break; } return 0; } void decrement(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { m->counts[idx]--; return; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; } } /* ============================================================ * Approaches * ============================================================ */ long long minRemoveTwoMaps(const int a[], int n, const int b[], int m) { FreqMap *ma = createMap(TABLE_SIZE); FreqMap *mb = createMap(TABLE_SIZE); if (ma == NULL || mb == NULL) { freeMap(ma); freeMap(mb); return -1; } for (int i = 0; i < n; i++) increment(ma, a[i]); for (int i = 0; i < m; i++) increment(mb, b[i]); long long res = 0; for (int i = 0; i < ma->capacity; i++) { if (!ma->occupied[i]) continue; int v = ma->keys[i]; int c2 = getCount(mb, v); if (c2 > 0) { int c1 = ma->counts[i]; res += (c1 < c2) ? c1 : c2; } } freeMap(ma); freeMap(mb); return res; } long long minRemoveSingleMap(const int a[], int n, const int b[], int m) { FreqMap *count = createMap(TABLE_SIZE); if (count == NULL) return -1; for (int i = 0; i < n; i++) increment(count, a[i]); long long res = 0; for (int i = 0; i < m; i++) { if (getCount(count, b[i]) > 0) { decrement(count, b[i]); res++; } } freeMap(count); return res; } /* ============================================================ * Utility * ============================================================ */ void printArray(const int arr[], int n) { printf("["); for (int i = 0; i < n; i++) { printf("%d", arr[i]); if (i < n - 1) printf(", "); } printf("]"); } static void demoBasic(void) { printf("\n=== Sample Arrays ===\n"); int a1[] = {2, 3, 4, 5, 8}; int b1[] = {1, 2, 3, 4}; int a2[] = {1, 2, 3, 4}; int b2[] = {5, 6, 7}; int a3[] = {1, 2, 2, 3}; int b3[] = {2, 2, 2, 4}; struct { int *a; int n; int *b; int m; const char *label; } tests[] = { {a1, 5, b1, 4, "arr1=[2,3,4,5,8], arr2=[1,2,3,4]"}, {a2, 4, b2, 3, "arr1=[1,2,3,4], arr2=[5,6,7]"}, {a3, 4, b3, 4, "arr1=[1,2,2,3], arr2=[2,2,2,4]"} }; int nt = sizeof(tests) / sizeof(tests[0]); for (int t = 0; t < nt; t++) { printf("\nTest %d: %s\n", t + 1, tests[t].label); printf(" arr1: "); printArray(tests[t].a, tests[t].n); printf("\n"); printf(" arr2: "); printArray(tests[t].b, tests[t].m); printf("\n"); long long r1 = minRemoveTwoMaps(tests[t].a, tests[t].n, tests[t].b, tests[t].m); long long r2 = minRemoveSingleMap(tests[t].a, tests[t].n, tests[t].b, tests[t].m); printf(" Two-map: %lld\n Single-map: %lld\n", r1, r2); printf(" Verify: %s\n", (r1 == r2) ? "PASS" : "FAIL"); } printf("\n"); } static void demoEdgeCases(void) { printf("\n=== Edge Cases ===\n"); /* Empty arrays */ printf("Both empty: %lld / %lld\n", minRemoveTwoMaps(NULL, 0, NULL, 0), minRemoveSingleMap(NULL, 0, NULL, 0)); /* One empty */ int a1[] = {1, 2, 3}; printf("arr2 empty: %lld / %lld\n", minRemoveTwoMaps(a1, 3, NULL, 0), minRemoveSingleMap(a1, 3, NULL, 0)); /* Disjoint */ int a2[] = {1, 2}; int b2[] = {3, 4}; printf("Disjoint: %lld / %lld (expect 0)\n", minRemoveTwoMaps(a2, 2, b2, 2), minRemoveSingleMap(a2, 2, b2, 2)); /* Identical */ int a3[] = {5, 5, 5}; int b3[] = {5, 5, 5}; printf("Identical [5,5,5]: %lld / %lld (expect 3)\n", minRemoveTwoMaps(a3, 3, b3, 3), minRemoveSingleMap(a3, 3, b3, 3)); /* Different frequencies */ int a4[] = {1, 1, 1, 2}; int b4[] = {1, 2, 2, 2}; printf("Different freqs: %lld / %lld (expect min(3,1)+min(1,3)=2)\n", minRemoveTwoMaps(a4, 4, b4, 4), minRemoveSingleMap(a4, 4, b4, 4)); /* Single common element */ int a5[] = {7, 8, 9}; int b5[] = {10, 7, 11}; printf("Single common: %lld / %lld (expect 1)\n", minRemoveTwoMaps(a5, 3, b5, 3), minRemoveSingleMap(a5, 3, b5, 3)); /* Negatives */ int a6[] = {-1, -2, -3}; int b6[] = {-2, -3, -4}; printf("Negatives: %lld / %lld (expect 2)\n", minRemoveTwoMaps(a6, 3, b6, 3), minRemoveSingleMap(a6, 3, b6, 3)); /* Zero */ int a7[] = {0, 0, 1}; int b7[] = {0, 2}; printf("Zero: %lld / %lld (expect min(2,1)=1)\n", minRemoveTwoMaps(a7, 3, b7, 2), minRemoveSingleMap(a7, 3, b7, 2)); } static void performanceTest(void) { printf("\n=== Performance Comparison ===\n"); int n = 5000000, m = 5000000; int *a = malloc((size_t)n * sizeof(int)); int *b = malloc((size_t)m * sizeof(int)); if (a == NULL || b == NULL) { free(a); free(b); return; } srand(42); for (int i = 0; i < n; i++) a[i] = rand() % 200000; for (int i = 0; i < m; i++) b[i] = rand() % 200000; clock_t start, end; double t; start = clock(); long long r1 = minRemoveTwoMaps(a, n, b, m); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Two-map (n=%d, m=%d): result=%lld, time=%.3f ms\n", n, m, r1, t); start = clock(); long long r2 = minRemoveSingleMap(a, n, b, m); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Single-map (n=%d, m=%d): result=%lld, time=%.3f ms\n", n, m, r2, t); printf("Verify: %s\n", (r1 == r2) ? "PASS" : "FAIL"); free(a); free(b); } static int readInteger(const char *prompt, int *outValue) { char buffer[MAX_LINE_LEN]; char *endPtr; long val; printf("%s", prompt); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) return -1; errno = 0; val = strtol(buffer, &endPtr, 10); if (endPtr == buffer || (*endPtr != '\n' && *endPtr != '\0')) return -1; if (errno == ERANGE || val < -2147483647 - 1 || val > 2147483647) return -1; *outValue = (int)val; return 0; } static void demoInteractive(void) { int n, m; if (readInteger("Enter arr1 size (0-10000): ", &n) != 0 || n < 0 || n > 10000) { printf("Invalid size.\n"); return; } int *a = NULL; if (n > 0) { a = malloc((size_t)n * sizeof(int)); if (a == NULL) return; for (int i = 0; i < n; i++) { char p[40]; snprintf(p, sizeof(p), " arr1[%d]: ", i); if (readInteger(p, &a[i]) != 0) { free(a); return; } } } if (readInteger("Enter arr2 size (0-10000): ", &m) != 0 || m < 0 || m > 10000) { printf("Invalid size.\n"); free(a); return; } int *b = NULL; if (m > 0) { b = malloc((size_t)m * sizeof(int)); if (b == NULL) { free(a); return; } for (int i = 0; i < m; i++) { char p[40]; snprintf(p, sizeof(p), " arr2[%d]: ", i); if (readInteger(p, &b[i]) != 0) { free(a); free(b); return; } } } printf("\nMinimum removals: %lld\n", minRemoveSingleMap(a, n, b, m)); free(a); free(b); } int main(void) { int choice; char buffer[MAX_LINE_LEN]; printf("=== Minimum Removals so No Common Elements Demo ===\n"); while (1) { printf("\n--- Menu ---\n"); printf("1. Demonstrate Both Approaches\n"); printf("2. Edge Cases\n"); printf("3. Performance Comparison\n"); printf("4. Interactive\n"); printf("5. Exit\n"); printf("Enter choice: "); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) { fprintf(stderr, "Error reading choice.\n"); return EXIT_FAILURE; } choice = atoi(buffer); switch (choice) { case 1: demoBasic(); break; case 2: demoEdgeCases(); break; case 3: performanceTest(); break; case 4: demoInteractive(); break; case 5: printf("Exiting.\n"); return EXIT_SUCCESS; default: printf("Invalid choice.\n"); break; } } return EXIT_SUCCESS; }

Explanation of the Approaches

  1. Two maps: Count frequencies of both arrays; for each shared value, add min(f1, f2).
  2. Single map: Count arr1; iterate arr2 and decrement each found value, adding 1 per match.

Both compute the same sum. The single map saves memory and cache pressure.

Expected Output (sample run)

text

=== Sample Arrays === Test 1: arr1=[2,3,4,5,8], arr2=[1,2,3,4] arr1: [2, 3, 4, 5, 8] arr2: [1, 2, 3, 4] Two-map: 3 Single-map: 3 Verify: PASS Test 2: arr1=[1,2,3,4], arr2=[5,6,7] arr1: [1, 2, 3, 4] arr2: [5, 6, 7] Two-map: 0 Single-map: 0 Verify: PASS Test 3: arr1=[1,2,2,3], arr2=[2,2,2,4] arr1: [1, 2, 2, 3] arr2: [2, 2, 2, 4] Two-map: 2 Single-map: 2 Verify: PASS === Edge Cases === Both empty: 0 / 0 arr2 empty: 0 / 0 Disjoint: 0 / 0 (expect 0) Identical [5,5,5]: 3 / 3 (expect 3) Different freqs: 2 / 2 (expect min(3,1)+min(1,3)=2) Single common: 1 / 1 (expect 1) Negatives: 2 / 2 (expect 2) Zero: 1 / 1 (expect min(2,1)=1)

Common Pitfalls, Edge Cases & Memory Considerations

1. Forgetting Frequencies

Two arrays with the same value but different counts need careful handling.

c

arr1 = [1, 1, 1] arr2 = [1] → remove 1 (min(3, 1) = 1)

2. Using Sets Instead of Maps

A set ignores frequencies and overcounts.

Solution: Use frequency maps.

3. Counting Per Value vs. Total

The answer sums min(f1, f2) across all shared values, not just one.

4. Integer Overflow in Sum

For large arrays with many duplicates, the sum can exceed int.

Solution: Use long long.

5. Empty Arrays

Any of n = 0 or m = 0 yields 0 (no removals needed).

6. Hash Table Sizing

Undersized tables cause probe failures.

Solution: Size ≥ 2× distinct values.

7. Negative Keys

Cast to unsigned int in the hash function.

8. Mutating Input

Both approaches leave inputs intact.

9. decrement Call on Missing Key

Guard with getCount() > 0 before decrementing.

10. Memory Considerations

  • Two maps: O(n + m) space.
  • Single map: O(n) space — preferred.
  • Free allocated memory.
  • Use long long for the total.

Summary & Best Practices

The minimum-removals problem reduces to summing min(f1(v), f2(v)) over all shared values. The single-map approach achieves O(n + m) time and O(n) space.

Key Takeaways

  • Answer = Σ min(f1(v), f2(v)) for v common to both arrays.
  • Two maps: O(n + m) time, O(n + m) space.
  • Single map: O(n + m) time, O(n) space — preferred.
  • Remove from the side with fewer copies per value.
  • Use long long for the total.

Best Practices

  1. Use the single-map approach for better memory usage.
  2. Track frequencies, not just membership.
  3. Sum min(f1, f2) per common value.
  4. Use long long for the answer.
  5. Guard decrements with a positive count check.
  6. Size hash tables ≥ 2× distinct values.
  7. Cast negative keys to unsigned int.
  8. Free all allocated memory.
  9. Test edge cases: empty, disjoint, identical, unequal frequencies.
  10. Profile for large inputs.

By mastering this problem, you’ll deepen your intuition for multiset operations, frequency counting, and set-theoretic reductions.

Learn to find minimum subsets with distinct elements in C: the answer equals the max frequency. Compare naive, sorted, and hash O(n) approaches.
C Programming

Minimum Subsets with Distinct Elements in C

Minimum Subsets with Distinct Elements in C: A Comprehensive Tutorial

Given an array of integers, partition its elements into the minimum number of subsets so that no subset contains duplicates. At first glance, this looks like a combinatorial optimization problem—but there’s an elegant insight that collapses it into a single statistic: the maximum frequency of any element.

This tutorial explores the reasoning, three implementation approaches (naive O(n²), sort O(n log n), hash O(n)), and everything you need to write robust C code that solves it.

Conceptual Introduction & Fundamentals

What is the Problem?

Given an array arr[], distribute its elements into the fewest possible subsets, where each subset must contain distinct elements (no duplicates inside a subset).

Examples:

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Input: arr = [1, 2, 3, 4] Output: 1 (all distinct → one subset) Input: arr = [1, 2, 3, 3] Output: 2 ({1,2,3} and {3})

Why This Problem Matters

  • Pattern recognition: Reduces a seemingly complex problem to one statistic.
  • Frequency counting: Foundation for many counting and partition problems.
  • Combinatorial reasoning: Pigeonhole principle in action.
  • Interview preparation: Common coding interview question.
  • Real-world applications: Task scheduling, resource allocation, exam seating.

The Key Insight

Every subset can contain at most one copy of any value. If a value v appears f(v) times, we need at least f(v) subsets to place all its copies. The most frequent value determines the minimum:

text

answer = max over all values of frequency

Why the Answer is Sufficient

Suppose the maximum frequency is m for value x. Create m empty subsets. Distribute the elements round-robin (each value goes to a different subset each time it appears). Since every value appears at most m times, each value fits into distinct subsets, so no subset gets a duplicate.

Approaches Overview

ApproachTimeSpace
Naive nested loopsO(n²)O(1)
SortingO(n log n)O(1) or O(n)
Hash mapO(n) averageO(n)

Syntax & Core Mechanics

Approach 1: Naive Nested Loops

For each element, count its occurrences by scanning the array.

c

#include <stdio.h> int minSubsetsNaive(const int arr[], int n) { if (n <= 0) return 0; int maxCount = 0; for (int i = 0; i < n; i++) { int count = 0; for (int j = 0; j < n; j++) { if (arr[i] == arr[j]) count++; } if (count > maxCount) maxCount = count; } return maxCount; } int main(void) { int arr[] = {40, 50, 30, 40, 50, 30, 30}; int n = sizeof(arr) / sizeof(arr[0]); printf("%d\n", minSubsetsNaive(arr, n)); return 0; }

Expected output:

text

3

Time: O(n²), Space: O(1)

Approach 2: Sort + Linear Scan

After sorting, equal elements are adjacent. Track the run length.

c

#include <stdio.h> #include <stdlib.h> #include <string.h> static int compareInts(const void *a, const void *b) { int x = *(const int *)a; int y = *(const int *)b; return (x > y) - (x < y); } int minSubsetsSorted(const int arr_in[], int n) { if (n <= 0) return 0; if (n == 1) return 1; int *arr = malloc((size_t)n * sizeof(int)); if (arr == NULL) return -1; memcpy(arr, arr_in, (size_t)n * sizeof(int)); qsort(arr, n, sizeof(int), compareInts); int maxCount = 1; int currCount = 1; for (int i = 1; i < n; i++) { if (arr[i] == arr[i - 1]) currCount++; else currCount = 1; if (currCount > maxCount) maxCount = currCount; } free(arr); return maxCount; } int main(void) { int arr[] = {40, 50, 30, 40, 50, 30, 30}; int n = sizeof(arr) / sizeof(arr[0]); printf("%d\n", minSubsetsSorted(arr, n)); return 0; }

Expected output:

text

3

Time: O(n log n), Space: O(n) for copy

Approach 3: Hash Map (Optimal)

Insert into a hash map (key = value, value = frequency). Return the maximum frequency.

c

#include <stdio.h> #include <stdlib.h> #define TABLE_SIZE 4096 typedef struct { int *keys; int *counts; int *occupied; int capacity; } FreqMap; static unsigned int hashFunc(int key, int cap) { return ((unsigned int)key) % (unsigned int)cap; } FreqMap *createMap(int cap) { FreqMap *m = malloc(sizeof(FreqMap)); if (m == NULL) return NULL; m->keys = malloc((size_t)cap * sizeof(int)); m->counts = calloc((size_t)cap, sizeof(int)); m->occupied = calloc((size_t)cap, sizeof(int)); if (m->keys == NULL || m->counts == NULL || m->occupied == NULL) { free(m->keys); free(m->counts); free(m->occupied); free(m); return NULL; } m->capacity = cap; return m; } void freeMap(FreqMap *m) { if (m) { free(m->keys); free(m->counts); free(m->occupied); free(m); } } void increment(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { m->counts[idx]++; return; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; } m->keys[idx] = key; m->counts[idx] = 1; m->occupied[idx] = 1; } int minSubsetsHash(const int arr[], int n) { if (n <= 0) return 0; FreqMap *m = createMap(TABLE_SIZE); if (m == NULL) return -1; for (int i = 0; i < n; i++) increment(m, arr[i]); int maxCount = 0; for (int i = 0; i < m->capacity; i++) { if (m->occupied[i] && m->counts[i] > maxCount) { maxCount = m->counts[i]; } } freeMap(m); return maxCount; } int main(void) { int arr[] = {40, 50, 30, 40, 50, 30, 30}; int n = sizeof(arr) / sizeof(arr[0]); printf("%d\n", minSubsetsHash(arr, n)); return 0; }

Expected output:

text

3

Time: O(n) average, Space: O(n)

Round-Robin Distribution Proof

If the maximum frequency is m for value x:

  1. Create m empty subsets.
  2. For each distinct value v appearing f(v) ≤ m times, assign its f(v) copies to f(v) distinct subsets.
  3. Each subset gets at most one copy of v.

This yields a valid partition with exactly m subsets, and no smaller count can work (pigeonhole).

Full Practical Code Example with Explanation

This program implements all three approaches and verifies they agree on multiple datasets.

c

/** * min_subsets.c * Minimum number of subsets with distinct elements. * Compares naive, sorted, and hash map approaches. */ #include <stdio.h> #include <stdlib.h> #include <string.h> #include <errno.h> #include <time.h> #define MAX_LINE_LEN 100 #define TABLE_SIZE 262144 /* ============================================================ * FreqMap * ============================================================ */ typedef struct { int *keys; int *counts; int *occupied; int capacity; } FreqMap; static unsigned int hashFunc(int key, int cap) { return ((unsigned int)key) % (unsigned int)cap; } FreqMap *createMap(int cap) { FreqMap *m = malloc(sizeof(FreqMap)); if (m == NULL) return NULL; m->keys = malloc((size_t)cap * sizeof(int)); m->counts = calloc((size_t)cap, sizeof(int)); m->occupied = calloc((size_t)cap, sizeof(int)); if (m->keys == NULL || m->counts == NULL || m->occupied == NULL) { free(m->keys); free(m->counts); free(m->occupied); free(m); return NULL; } m->capacity = cap; return m; } void freeMap(FreqMap *m) { if (m) { free(m->keys); free(m->counts); free(m->occupied); free(m); } } void increment(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { m->counts[idx]++; return; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; } m->keys[idx] = key; m->counts[idx] = 1; m->occupied[idx] = 1; } /* ============================================================ * Approaches * ============================================================ */ int minSubsetsNaive(const int arr[], int n) { if (n <= 0) return 0; int maxCount = 0; for (int i = 0; i < n; i++) { int cnt = 0; for (int j = 0; j < n; j++) if (arr[i] == arr[j]) cnt++; if (cnt > maxCount) maxCount = cnt; } return maxCount; } static int compareInts(const void *a, const void *b) { int x = *(const int *)a; int y = *(const int *)b; return (x > y) - (x < y); } int minSubsetsSorted(const int arr_in[], int n) { if (n <= 0) return 0; if (n == 1) return 1; int *arr = malloc((size_t)n * sizeof(int)); if (arr == NULL) return -1; memcpy(arr, arr_in, (size_t)n * sizeof(int)); qsort(arr, n, sizeof(int), compareInts); int maxCount = 1, currCount = 1; for (int i = 1; i < n; i++) { if (arr[i] == arr[i - 1]) currCount++; else currCount = 1; if (currCount > maxCount) maxCount = currCount; } free(arr); return maxCount; } int minSubsetsHash(const int arr[], int n) { if (n <= 0) return 0; FreqMap *m = createMap(TABLE_SIZE); if (m == NULL) return -1; for (int i = 0; i < n; i++) increment(m, arr[i]); int maxCount = 0; for (int i = 0; i < m->capacity; i++) { if (m->occupied[i] && m->counts[i] > maxCount) { maxCount = m->counts[i]; } } freeMap(m); return maxCount; } /* ============================================================ * Utility * ============================================================ */ void printArray(const int arr[], int n) { printf("["); for (int i = 0; i < n; i++) { printf("%d", arr[i]); if (i < n - 1) printf(", "); } printf("]"); } static void demoBasic(void) { printf("\n=== Sample Arrays ===\n"); int a1[] = {1, 2, 3, 4}; int a2[] = {1, 2, 3, 3}; int a3[] = {40, 50, 30, 40, 50, 30, 30}; int a4[] = {1, 2, 1, 2, 3, 3, 2, 2}; struct { int *arr; int n; const char *label; } tests[] = { {a1, 4, "[1,2,3,4]"}, {a2, 4, "[1,2,3,3]"}, {a3, 7, "[40,50,30,40,50,30,30]"}, {a4, 8, "[1,2,1,2,3,3,2,2]"} }; int nt = sizeof(tests) / sizeof(tests[0]); for (int t = 0; t < nt; t++) { printf("\nTest %d: %s\n", t + 1, tests[t].label); printf(" Array: "); printArray(tests[t].arr, tests[t].n); printf("\n"); int r1 = minSubsetsNaive(tests[t].arr, tests[t].n); int r2 = minSubsetsSorted(tests[t].arr, tests[t].n); int r3 = minSubsetsHash(tests[t].arr, tests[t].n); printf(" Naive: %d\n Sorted: %d\n Hash: %d\n", r1, r2, r3); printf(" Verify: %s\n", (r1 == r2 && r2 == r3) ? "PASS" : "FAIL"); } printf("\n"); } static void demoEdgeCases(void) { printf("\n=== Edge Cases ===\n"); /* Empty */ printf("Empty array: naive=%d, sorted=%d, hash=%d\n", minSubsetsNaive(NULL, 0), minSubsetsSorted(NULL, 0), minSubsetsHash(NULL, 0)); /* Single */ int a1[] = {7}; printf("Single element: %d / %d / %d\n", minSubsetsNaive(a1, 1), minSubsetsSorted(a1, 1), minSubsetsHash(a1, 1)); /* All distinct */ int a2[] = {1, 2, 3, 4, 5}; printf("All distinct: %d / %d / %d (expect 1)\n", minSubsetsNaive(a2, 5), minSubsetsSorted(a2, 5), minSubsetsHash(a2, 5)); /* All same */ int a3[] = {9, 9, 9, 9}; printf("All same: %d / %d / %d (expect 4)\n", minSubsetsNaive(a3, 4), minSubsetsSorted(a3, 4), minSubsetsHash(a3, 4)); /* One dominant value */ int a4[] = {1, 1, 1, 1, 2, 3}; printf("One dominant: %d / %d / %d (expect 4)\n", minSubsetsNaive(a4, 6), minSubsetsSorted(a4, 6), minSubsetsHash(a4, 6)); /* Negatives */ int a5[] = {-1, -1, -2, -2, -3}; printf("Negatives: %d / %d / %d (expect 2)\n", minSubsetsNaive(a5, 5), minSubsetsSorted(a5, 5), minSubsetsHash(a5, 5)); /* Two values alternating */ int a6[] = {5, 6, 5, 6, 5, 6}; printf("Alternating: %d / %d / %d (expect 3)\n", minSubsetsNaive(a6, 6), minSubsetsSorted(a6, 6), minSubsetsHash(a6, 6)); /* Zeros */ int a7[] = {0, 0, 0, 1, 1}; printf("Zeros: %d / %d / %d (expect 3)\n", minSubsetsNaive(a7, 5), minSubsetsSorted(a7, 5), minSubsetsHash(a7, 5)); } static void performanceTest(void) { printf("\n=== Performance Comparison ===\n"); /* Naive small */ int smallN = 3000; int *sa = malloc((size_t)smallN * sizeof(int)); if (sa == NULL) return; srand(42); for (int i = 0; i < smallN; i++) sa[i] = rand() % 500; clock_t start, end; double t; start = clock(); int r1 = minSubsetsNaive(sa, smallN); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Naive (n=%d): result=%d, time=%.3f ms\n", smallN, r1, t); start = clock(); int r2 = minSubsetsSorted(sa, smallN); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Sorted (n=%d): result=%d, time=%.3f ms\n", smallN, r2, t); start = clock(); int r3 = minSubsetsHash(sa, smallN); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Hash (n=%d): result=%d, time=%.3f ms\n", smallN, r3, t); printf("Verify: %s\n", (r1 == r2 && r2 == r3) ? "PASS" : "FAIL"); free(sa); /* Large hash */ int largeN = 5000000; int *la = malloc((size_t)largeN * sizeof(int)); if (la == NULL) return; for (int i = 0; i < largeN; i++) la[i] = rand() % 100000; start = clock(); int r4 = minSubsetsHash(la, largeN); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Hash (n=%d): result=%d, time=%.3f ms\n", largeN, r4, t); free(la); } static int readInteger(const char *prompt, int *outValue) { char buffer[MAX_LINE_LEN]; char *endPtr; long val; printf("%s", prompt); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) return -1; errno = 0; val = strtol(buffer, &endPtr, 10); if (endPtr == buffer || (*endPtr != '\n' && *endPtr != '\0')) return -1; if (errno == ERANGE || val < -2147483647 - 1 || val > 2147483647) return -1; *outValue = (int)val; return 0; } static void demoInteractive(void) { int n; if (readInteger("Enter array size (0-10000): ", &n) != 0 || n < 0 || n > 10000) { printf("Invalid size.\n"); return; } int *arr = NULL; if (n > 0) { arr = malloc((size_t)n * sizeof(int)); if (arr == NULL) return; for (int i = 0; i < n; i++) { char p[40]; snprintf(p, sizeof(p), " arr[%d]: ", i); if (readInteger(p, &arr[i]) != 0) { printf("Invalid value.\n"); free(arr); return; } } } printf("\nMinimum subsets: %d\n", minSubsetsHash(arr, n)); free(arr); } int main(void) { int choice; char buffer[MAX_LINE_LEN]; printf("=== Minimum Subsets with Distinct Elements Demo ===\n"); while (1) { printf("\n--- Menu ---\n"); printf("1. Demonstrate All Approaches\n"); printf("2. Edge Cases\n"); printf("3. Performance Comparison\n"); printf("4. Interactive\n"); printf("5. Exit\n"); printf("Enter choice: "); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) { fprintf(stderr, "Error reading choice.\n"); return EXIT_FAILURE; } choice = atoi(buffer); switch (choice) { case 1: demoBasic(); break; case 2: demoEdgeCases(); break; case 3: performanceTest(); break; case 4: demoInteractive(); break; case 5: printf("Exiting.\n"); return EXIT_SUCCESS; default: printf("Invalid choice.\n"); break; } } return EXIT_SUCCESS; }

Explanation of the Approaches

  1. Naive: For each element, count its frequency with a nested loop; track the maximum. O(n²).
  2. Sorted: Sort; scan runs of equal elements. O(n log n).
  3. Hash: Increment frequency in a hash map; find the maximum. O(n) average.

Expected Output (sample run)

text

=== Sample Arrays === Test 1: [1,2,3,4] Array: [1, 2, 3, 4] Naive: 1 Sorted: 1 Hash: 1 Verify: PASS Test 2: [1,2,3,3] Array: [1, 2, 3, 3] Naive: 2 Sorted: 2 Hash: 2 Verify: PASS Test 3: [40,50,30,40,50,30,30] Array: [40, 50, 30, 40, 50, 30, 30] Naive: 3 Sorted: 3 Hash: 3 Verify: PASS Test 4: [1,2,1,2,3,3,2,2] Array: [1, 2, 1, 2, 3, 3, 2, 2] Naive: 4 Sorted: 4 Hash: 4 Verify: PASS === Edge Cases === Empty array: naive=0, sorted=0, hash=0 Single element: 1 / 1 / 1 All distinct: 1 / 1 / 1 (expect 1) All same: 4 / 4 / 4 (expect 4) One dominant: 4 / 4 / 4 (expect 4) Negatives: 2 / 2 / 2 (expect 2) Alternating: 3 / 3 / 3 (expect 3) Zeros: 3 / 3 / 3 (expect 3)

Common Pitfalls, Edge Cases & Memory Considerations

1. Confusing “Distinct” with “Unique Values”

Each subset must have distinct elements within itself; elements may repeat across subsets.

2. Empty Array

Return 0 (no subsets needed). Guard explicitly.

3. Single Element

Exactly 1 subset is required.

4. All Same Values

The answer equals n (each subset holds one copy).

5. All Distinct

The answer is 1 (one subset holds everything).

6. Large Frequencies

The count fits in int for arrays up to INT_MAX elements; use size_t for very large arrays.

7. Hash Table Sizing

Undersized table → probe failures.

Solution: Size ≥ 2× number of distinct values (or ≥ 2× n).

8. Negative Keys

Cast to unsigned int in the hash function.

9. Sorting Mutates Input

Copy before sorting if preservation needed.

10. Memory Considerations

  • Naive: O(1) extra space.
  • Sorted: O(n) for the copy.
  • Hash: O(n) for the frequency map.
  • Free all dynamic allocations.

Summary & Best Practices

The minimum-subsets problem is a beautiful case of “reduce complexity to one statistic.” The answer is simply the maximum frequency of any value in the array.

Key Takeaways

  • Answer = maximum frequency of any element.
  • All distinct → 1 subset.
  • All same → n subsets.
  • Naive: O(n²) — educational.
  • Sorted: O(n log n) — memory-friendly.
  • Hash: O(n) average — optimal.

Best Practices

  1. Use the hash map for O(n) average.
  2. Use sorting when memory is tight.
  3. Track the maximum frequency during the scan.
  4. Handle empty arrays (0) and single elements (1).
  5. Size hash tables ≥ 2× distinct values.
  6. Cast negative keys to unsigned int.
  7. Copy inputs before sorting if preservation matters.
  8. Free allocated memory.
  9. Test edge cases: empty, single, all-same, all-distinct, negatives.
  10. Profile for large inputs.

By mastering this problem, you’ll recognize reductions from combinatorial problems to single statistics—a powerful pattern across algorithms.

Learn to count pairs with absolute difference k in C: naive O(n²), sort+two pointers O(n log n), and hash map O(n) approaches with code examples.
C Programming

Count Pairs with Absolute Difference K in C

Count All Pairs with Absolute Difference Equal to K in C: A Comprehensive Tutorial

Counting pairs with an absolute difference equal to a target k is a classic problem that combines array traversal, sorting, and hashing. It demonstrates the same set of algorithmic trade-offs as Two Sum but requires careful handling of positive and negative differences — since the absolute value can come from either direction.

This tutorial provides an in-depth exploration of three approaches in C: naive O(n²), sort + two pointers O(n log n), and hash map O(n). Each implementation uses portable C99/C11 code with robust error handling and complete output examples.

Conceptual Introduction & Fundamentals

What is the Problem?

Given an array arr[] and a positive integer k, count pairs (i, j) with i < j such that |arr[i] - arr[j]| == k.

Examples:

text

Input: arr = [1, 4, 1, 4, 5], k = 3 Output: 4 Pairs: (1,4), (1,4), (1,4), (4,1) Input: arr = [8, 16, 12, 16, 4, 0], k = 4 Output: 5 Pairs: (8,12), (8,4), (16,12), (12,16), (4,0)

Why This Problem Matters

  • Absolute value handling: Two complementary lookups per element.
  • Combinatorial counting: Same-value pairs occur when k = 0.
  • Hash map mastery: Frequency counting with two-pronged complement lookup.
  • Algorithm comparison: O(n²) → O(n log n) → O(n).
  • Interview preparation: Common coding interview question.

Key Insight

For each arr[i], a valid partner x satisfies either:

  • x = arr[i] + k
  • x = arr[i] - k

Checking both possibilities against a frequency map counts all pairs with absolute difference k in one pass.

Approaches Overview

ApproachTimeSpace
Naive nested loopsO(n²)O(1)
Sort + two pointersO(n log n)O(1) or O(n)
Hash mapO(n) averageO(n)

Same-Element Case (k = 0)

When k = 0, both complements coincide. The correct count is C(freq, 2) = freq × (freq - 1) / 2. The hash approach handles this naturally: for the second occurrence, only the +0 lookup succeeds (since -0 is the same key), so each pair is counted exactly once.

Syntax & Core Mechanics

Approach 1: Naive Nested Loops

Enumerate all pairs; test absolute difference.

c

#include <stdio.h> #include <stdlib.h> long long countPairsNaive(const int arr[], int n, int k) { long long cnt = 0; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { long long diff = (long long)arr[i] - (long long)arr[j]; if (diff < 0) diff = -diff; if (diff == (long long)k) cnt++; } } return cnt; } int main(void) { int arr[] = {1, 4, 1, 4, 5}; int n = sizeof(arr) / sizeof(arr[0]); int k = 3; printf("%lld\n", countPairsNaive(arr, n, k)); return 0; }

Expected output:

text

4

Time: O(n²), Space: O(1)

Approach 2: Sort + Two Pointers

Sort; advance i and j so that the difference matches k. Handle equal-value groups combinatorially.

c

#include <stdio.h> #include <stdlib.h> static int compareInts(const void *a, const void *b) { int x = *(const int *)a; int y = *(const int *)b; return (x > y) - (x < y); } long long countPairsSorted(const int arr_in[], int n, int k) { if (n < 2) return 0; int *arr = malloc((size_t)n * sizeof(int)); if (arr == NULL) return -1; for (int i = 0; i < n; i++) arr[i] = arr_in[i]; qsort(arr, n, sizeof(int), compareInts); long long cnt = 0; int i = 0, j = 0; while (j < n) { long long diff = (long long)arr[j] - (long long)arr[i]; if (diff < k) { j++; } else if (diff > k) { i++; if (i > j) j = i; /* Keep j ≥ i */ } else { int ele1 = arr[i], ele2 = arr[j]; long long cnt1 = 0, cnt2 = 0; while (i < n && arr[i] == ele1) { i++; cnt1++; } while (j < n && arr[j] == ele2) { j++; cnt2++; } if (ele1 == ele2) cnt += cnt1 * (cnt1 - 1) / 2; else cnt += cnt1 * cnt2; } } free(arr); return cnt; } int main(void) { int arr[] = {1, 4, 1, 4, 5}; int n = sizeof(arr) / sizeof(arr[0]); int k = 3; printf("%lld\n", countPairsSorted(arr, n, k)); return 0; }

Expected output:

text

4

Time: O(n log n), Space: O(n) for copy

Approach 3: Hash Map (Optimal)

For each element, look up both arr[i] + k and arr[i] - k, then insert.

c

#include <stdio.h> #include <stdlib.h> #define TABLE_SIZE 8192 typedef struct { int *keys; long long *counts; int *occupied; int capacity; } FreqMap; static unsigned int hashFunc(int key, int cap) { return ((unsigned int)key) % (unsigned int)cap; } FreqMap *createMap(int cap) { FreqMap *m = malloc(sizeof(FreqMap)); if (m == NULL) return NULL; m->keys = malloc((size_t)cap * sizeof(int)); m->counts = calloc((size_t)cap, sizeof(long long)); m->occupied = calloc((size_t)cap, sizeof(int)); if (m->keys == NULL || m->counts == NULL || m->occupied == NULL) { free(m->keys); free(m->counts); free(m->occupied); free(m); return NULL; } m->capacity = cap; return m; } void freeMap(FreqMap *m) { if (m) { free(m->keys); free(m->counts); free(m->occupied); free(m); } } long long mapGet(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) return m->counts[idx]; idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) break; } return 0; } void mapIncrement(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { m->counts[idx]++; return; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; } m->keys[idx] = key; m->counts[idx] = 1; m->occupied[idx] = 1; } long long countPairsHash(const int arr[], int n, int k) { if (n < 2) return 0; FreqMap *m = createMap(TABLE_SIZE); if (m == NULL) return -1; long long cnt = 0; for (int i = 0; i < n; i++) { /* Partner greater by k */ cnt += mapGet(m, arr[i] + k); /* Partner smaller by k */ cnt += mapGet(m, arr[i] - k); /* Now record current element */ mapIncrement(m, arr[i]); } freeMap(m); return cnt; } int main(void) { int arr[] = {1, 4, 1, 4, 5}; int n = sizeof(arr) / sizeof(arr[0]); int k = 3; printf("%lld\n", countPairsHash(arr, n, k)); return 0; }

Expected output:

text

4

Time: O(n) average, Space: O(n)

Handling k = 0

When k == 0, both lookups are identical to arr[i]. The frequency counts prior occurrences, yielding exactly C(freq, 2) automatically. No special-casing required.

Full Practical Code Example with Explanation

This program compares all three approaches and exercises edge cases including k = 0, negatives, and large datasets.

c

/** * count_pairs_diff_k.c * Counts pairs with |arr[i] - arr[j]| == k. * Compares naive, sort+two-pointer, and hash map approaches. */ #include <stdio.h> #include <stdlib.h> #include <string.h> #include <errno.h> #include <time.h> #define MAX_LINE_LEN 100 #define TABLE_SIZE 262144 /* ============================================================ * FreqMap * ============================================================ */ typedef struct { int *keys; long long *counts; int *occupied; int capacity; } FreqMap; static unsigned int hashFunc(int key, int cap) { return ((unsigned int)key) % (unsigned int)cap; } FreqMap *createMap(int cap) { FreqMap *m = malloc(sizeof(FreqMap)); if (m == NULL) return NULL; m->keys = malloc((size_t)cap * sizeof(int)); m->counts = calloc((size_t)cap, sizeof(long long)); m->occupied = calloc((size_t)cap, sizeof(int)); if (m->keys == NULL || m->counts == NULL || m->occupied == NULL) { free(m->keys); free(m->counts); free(m->occupied); free(m); return NULL; } m->capacity = cap; return m; } void freeMap(FreqMap *m) { if (m) { free(m->keys); free(m->counts); free(m->occupied); free(m); } } long long mapGet(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) return m->counts[idx]; idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) break; } return 0; } void mapIncrement(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { m->counts[idx]++; return; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; } m->keys[idx] = key; m->counts[idx] = 1; m->occupied[idx] = 1; } /* ============================================================ * Approaches * ============================================================ */ long long countPairsNaive(const int arr[], int n, int k) { long long cnt = 0; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { long long diff = (long long)arr[i] - (long long)arr[j]; if (diff < 0) diff = -diff; if (diff == (long long)k) cnt++; } } return cnt; } static int compareInts(const void *a, const void *b) { int x = *(const int *)a; int y = *(const int *)b; return (x > y) - (x < y); } long long countPairsSorted(const int arr_in[], int n, int k) { if (n < 2) return 0; int *arr = malloc((size_t)n * sizeof(int)); if (arr == NULL) return -1; memcpy(arr, arr_in, (size_t)n * sizeof(int)); qsort(arr, n, sizeof(int), compareInts); long long cnt = 0; int i = 0, j = 0; while (j < n) { long long diff = (long long)arr[j] - (long long)arr[i]; if (diff < k) { j++; } else if (diff > k) { i++; if (i > j) j = i; } else { int ele1 = arr[i], ele2 = arr[j]; long long cnt1 = 0, cnt2 = 0; while (i < n && arr[i] == ele1) { i++; cnt1++; } while (j < n && arr[j] == ele2) { j++; cnt2++; } if (ele1 == ele2) cnt += cnt1 * (cnt1 - 1) / 2; else cnt += cnt1 * cnt2; } } free(arr); return cnt; } long long countPairsHash(const int arr[], int n, int k) { if (n < 2) return 0; FreqMap *m = createMap(TABLE_SIZE); if (m == NULL) return -1; long long cnt = 0; for (int i = 0; i < n; i++) { cnt += mapGet(m, arr[i] + k); cnt += mapGet(m, arr[i] - k); mapIncrement(m, arr[i]); } freeMap(m); return cnt; } /* ============================================================ * Utility * ============================================================ */ void printArray(const int arr[], int n) { printf("["); for (int i = 0; i < n; i++) { printf("%d", arr[i]); if (i < n - 1) printf(", "); } printf("]"); } static void demoBasic(void) { printf("\n=== Sample Arrays ===\n"); int a1[] = {1, 4, 1, 4, 5}; int a2[] = {8, 16, 12, 16, 4, 0}; struct { int *arr; int n; int k; const char *label; } tests[] = { {a1, 5, 3, "arr=[1,4,1,4,5], k=3"}, {a2, 6, 4, "arr=[8,16,12,16,4,0], k=4"} }; int nt = sizeof(tests) / sizeof(tests[0]); for (int t = 0; t < nt; t++) { printf("\nTest %d: %s\n", t + 1, tests[t].label); printf(" Array: "); printArray(tests[t].arr, tests[t].n); printf("\n"); long long r1 = countPairsNaive(tests[t].arr, tests[t].n, tests[t].k); long long r2 = countPairsSorted(tests[t].arr, tests[t].n, tests[t].k); long long r3 = countPairsHash(tests[t].arr, tests[t].n, tests[t].k); printf(" Naive: %lld\n Sorted: %lld\n Hash: %lld\n", r1, r2, r3); printf(" Verify: %s\n", (r1 == r2 && r2 == r3) ? "PASS" : "FAIL"); } printf("\n"); } static void demoEdgeCases(void) { printf("\n=== Edge Cases ===\n"); /* Empty */ printf("Empty array: naive=%lld, sorted=%lld, hash=%lld\n", countPairsNaive(NULL, 0, 1), countPairsSorted(NULL, 0, 1), countPairsHash(NULL, 0, 1)); /* Single */ int a1[] = {42}; printf("Single element: %lld / %lld / %lld\n", countPairsNaive(a1, 1, 5), countPairsSorted(a1, 1, 5), countPairsHash(a1, 1, 5)); /* Two equal, k=0 */ int a2[] = {5, 5}; printf("[5,5] k=0: %lld / %lld / %lld (expect 1)\n", countPairsNaive(a2, 2, 0), countPairsSorted(a2, 2, 0), countPairsHash(a2, 2, 0)); /* All same, k=0, freq=6 → C(6,2)=15 */ int a3[] = {7, 7, 7, 7, 7, 7}; printf("[7×6] k=0: %lld / %lld / %lld (expect 15)\n", countPairsNaive(a3, 6, 0), countPairsSorted(a3, 6, 0), countPairsHash(a3, 6, 0)); /* No pairs */ int a4[] = {1, 2, 3}; printf("[1,2,3] k=10: %lld / %lld / %lld\n", countPairsNaive(a4, 3, 10), countPairsSorted(a4, 3, 10), countPairsHash(a4, 3, 10)); /* Negatives */ int a5[] = {-1, -4, -1, -4, -5}; printf("[-1,-4,-1,-4,-5] k=3: %lld / %lld / %lld\n", countPairsNaive(a5, 5, 3), countPairsSorted(a5, 5, 3), countPairsHash(a5, 5, 3)); /* Zero and positives */ int a6[] = {0, 3, 0, 3}; printf("[0,3,0,3] k=3: %lld / %lld / %lld (expect 4)\n", countPairsNaive(a6, 4, 3), countPairsSorted(a6, 4, 3), countPairsHash(a6, 4, 3)); /* Large values, k small */ int a7[] = {1000000000, 1000000001, 999999999}; printf("Large values k=1: %lld / %lld / %lld\n", countPairsNaive(a7, 3, 1), countPairsSorted(a7, 3, 1), countPairsHash(a7, 3, 1)); } static void performanceTest(void) { printf("\n=== Performance Comparison ===\n"); /* Naive small */ int smallN = 3000; int *sa = malloc((size_t)smallN * sizeof(int)); if (sa == NULL) return; srand(42); for (int i = 0; i < smallN; i++) sa[i] = (rand() % 2000) - 1000; int k = 500; clock_t start, end; double t; start = clock(); long long r1 = countPairsNaive(sa, smallN, k); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Naive (n=%d): result=%lld, time=%.3f ms\n", smallN, r1, t); start = clock(); long long r2 = countPairsSorted(sa, smallN, k); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Sorted (n=%d): result=%lld, time=%.3f ms\n", smallN, r2, t); start = clock(); long long r3 = countPairsHash(sa, smallN, k); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Hash (n=%d): result=%lld, time=%.3f ms\n", smallN, r3, t); printf("Verify: %s\n", (r1 == r2 && r2 == r3) ? "PASS" : "FAIL"); free(sa); /* Large hash */ int largeN = 5000000; int *la = malloc((size_t)largeN * sizeof(int)); if (la == NULL) return; for (int i = 0; i < largeN; i++) la[i] = rand() % 100000; start = clock(); long long r4 = countPairsHash(la, largeN, 500); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Hash (n=%d, k=500): result=%lld, time=%.3f ms\n", largeN, r4, t); free(la); } static int readInteger(const char *prompt, int *outValue) { char buffer[MAX_LINE_LEN]; char *endPtr; long val; printf("%s", prompt); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) return -1; errno = 0; val = strtol(buffer, &endPtr, 10); if (endPtr == buffer || (*endPtr != '\n' && *endPtr != '\0')) return -1; if (errno == ERANGE || val < -2147483647 - 1 || val > 2147483647) return -1; *outValue = (int)val; return 0; } static void demoInteractive(void) { int n, k; if (readInteger("Enter array size (1-10000): ", &n) != 0 || n < 1 || n > 10000) { printf("Invalid size.\n"); return; } int *arr = malloc((size_t)n * sizeof(int)); if (arr == NULL) return; for (int i = 0; i < n; i++) { char prompt[40]; snprintf(prompt, sizeof(prompt), " arr[%d]: ", i); if (readInteger(prompt, &arr[i]) != 0) { printf("Invalid value.\n"); free(arr); return; } } if (readInteger("Enter k (non-negative): ", &k) != 0 || k < 0) { printf("Invalid k.\n"); free(arr); return; } printf("\nArray: "); printArray(arr, n); printf("\n"); printf("Pair count: %lld\n", countPairsHash(arr, n, k)); free(arr); } int main(void) { int choice; char buffer[MAX_LINE_LEN]; printf("=== Count Pairs with Absolute Difference K Demo ===\n"); while (1) { printf("\n--- Menu ---\n"); printf("1. Demonstrate All Approaches\n"); printf("2. Edge Cases\n"); printf("3. Performance Comparison\n"); printf("4. Interactive\n"); printf("5. Exit\n"); printf("Enter choice: "); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) { fprintf(stderr, "Error reading choice.\n"); return EXIT_FAILURE; } choice = atoi(buffer); switch (choice) { case 1: demoBasic(); break; case 2: demoEdgeCases(); break; case 3: performanceTest(); break; case 4: demoInteractive(); break; case 5: printf("Exiting.\n"); return EXIT_SUCCESS; default: printf("Invalid choice.\n"); break; } } return EXIT_SUCCESS; }

Explanation of the Approaches

  1. Naive: Compute |arr[i] - arr[j]| for all pairs. Simple, quadratic.
  2. Sorted + Two Pointers: After sorting, use sliding pointers so the difference is monotone.
  3. Hash Map: For each element, look up arr[i] + k and arr[i] - k before insertion.

Expected Output (sample run)

text

=== Sample Arrays === Test 1: arr=[1,4,1,4,5], k=3 Array: [1, 4, 1, 4, 5] Naive: 4 Sorted: 4 Hash: 4 Verify: PASS Test 2: arr=[8,16,12,16,4,0], k=4 Array: [8, 16, 12, 16, 4, 0] Naive: 5 Sorted: 5 Hash: 5 Verify: PASS === Edge Cases === Empty array: naive=0, sorted=0, hash=0 Single element: 0 / 0 / 0 [5,5] k=0: 1 / 1 / 1 (expect 1) [7×6] k=0: 15 / 15 / 15 (expect 15) [1,2,3] k=10: 0 / 0 / 0 [-1,-4,-1,-4,-5] k=3: 4 / 4 / 4 [0,3,0,3] k=3: 4 / 4 / 4 (expect 4) Large values k=1: 2 / 2 / 2

Common Pitfalls, Edge Cases & Memory Considerations

1. int Overflow in the Count

For large arrays with many duplicates, the count may exceed int.

Solution: Use long long for the count.

2. Integer Overflow in Difference

arr[i] - arr[j] can overflow int.

Solution: Cast operands to long long before subtracting.

3. Double Counting in Hash Map

Checking +k and -k looks redundant, but each finds a different pair orientation. Only pairs where the current element is the larger (via -k) or smaller (via +k) side are counted once each.

4. k = 0 Handling

Both lookups target arr[i]. The frequency counts prior occurrences; each pair is counted once, yielding C(freq, 2) automatically.

5. Negative k

The problem specifies k is positive; still guard against negative input.

6. Hash Table Sizing

Undersized table → probe failures.

Solution: Size ≥ 2× n.

7. Negatives Keys

Cast to unsigned int in the hash function.

8. Two-Pointer Invariant

Keep j ≥ i when moving i.

c

if (i > j) j = i;

9. Mutating Inputs in Sort

Copy before sorting if preservation is needed.

10. Memory Considerations

  • Naive: O(1).
  • Sorted: O(n) for the copy.
  • Hash: O(n) for the frequency map.
  • Free all dynamic allocations.

Summary & Best Practices

Counting pairs with absolute difference k is a classic problem with three clean solutions. The hash-map approach is optimal at O(n) average time using two-pronged complement lookup.

Key Takeaways

  • Pair condition: |arr[i] - arr[j]| == k.
  • Naive: O(n²).
  • Sort + two pointers: O(n log n), combinatorial counting for equal groups.
  • Hash map: O(n) average with two complement lookups per element.
  • k = 0 handled naturally by frequency counting.
  • Use long long for sums and counts.

Best Practices

  1. Use the hash map for O(n) average performance.
  2. Use long long for the count and differences.
  3. Look up both arr[i] + k and arr[i] - k before insertion.
  4. Cast negative keys to unsigned int.
  5. Size hash tables ≥ 2× n.
  6. Handle k = 0 via natural frequency counting.
  7. Keep j ≥ i in the sorted two-pointer approach.
  8. Free allocated memory.
  9. Test edge cases: empty, single, k = 0, negatives, large values.
  10. Profile for large inputs.

By mastering this problem, you’ll be equipped for count-based pair problems involving absolute value, distance, and metric-based lookups.

Learn to count pairs with a given sum in C: naive, sort+two pointers, and hash map O(n) approaches with combinatorial counting and edge cases.
C Programming

Count Pairs With Given Sum in C: Complete Guide

2 Sum – Count Pairs with Given Sum in C: A Comprehensive Tutorial

Counting the number of pairs in an array whose sum equals a target value is a classic problem that extends the well-known Two Sum. Instead of merely checking whether a pair exists, we must count all valid pairs, including cases where elements repeat.

This tutorial provides an in-depth exploration of three approaches in C: naive O(n²), sort + two pointers O(n log n), and hash map O(n). Each is implemented with portable C99/C11 code, robust error handling, and complete output examples.

Conceptual Introduction & Fundamentals

What is the Problem?

Given an integer array arr[] and a target value, count the number of distinct index pairs (i, j) with i < j such that arr[i] + arr[j] == target.

Examples:

text

Input: arr = [1, 5, 7, -1, 5], target = 6 Output: 3 Explanation: (1,5), (7,-1), (1,5) Input: arr = [1, 1, 1, 1], target = 2 Output: 6 Explanation: 4C2 = 6 unordered pairs of identical elements Input: arr = [10, 12, 10, 15, -1], target = 125 Output: 0

Why This Problem Matters

  • Counting combinatorics: Same-value pairs require the formula C(k, 2) = k(k-1)/2.
  • Hash map mastery: Frequency-based counting in one pass.
  • Algorithm comparison: O(n²) → O(n log n) → O(n).
  • Real-world relevance: Analytics, matching systems, and graph edge counting.
  • Interview preparation: Common coding interview question.

Approaches Overview

ApproachTimeSpace
Naive nested loopsO(n²)O(1)
Sort + two pointersO(n log n)O(1) or O(n)
Hash mapO(n) averageO(n)

The Same-Element Formula

When both elements are equal (2 × value == target), the number of pairs is C(freq, 2) = freq × (freq - 1) / 2.

Syntax & Core Mechanics

Approach 1: Naive Nested Loops

Enumerate all pairs (i, j) with i < j.

c

#include <stdio.h> int countPairsNaive(const int arr[], int n, int target) { int cnt = 0; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { if (arr[i] + arr[j] == target) cnt++; } } return cnt; } int main(void) { int arr[] = {1, 5, 7, -1, 5}; int n = sizeof(arr) / sizeof(arr[0]); int target = 6; printf("%d\n", countPairsNaive(arr, n, target)); return 0; }

Expected output:

text

3

Time: O(n²), Space: O(1)

Approach 2: Sort + Two Pointers

Sort the array; move left/right inward; handle groups of equal values.

c

#include <stdio.h> #include <stdlib.h> static int compareInts(const void *a, const void *b) { int x = *(const int *)a; int y = *(const int *)b; return (x > y) - (x < y); } long long countPairsSorted(const int arr_in[], int n, int target) { if (n < 2) return 0; int *arr = malloc((size_t)n * sizeof(int)); if (arr == NULL) return -1; for (int i = 0; i < n; i++) arr[i] = arr_in[i]; qsort(arr, n, sizeof(int), compareInts); long long res = 0; int left = 0, right = n - 1; while (left < right) { int sum = arr[left] + arr[right]; if (sum < target) { left++; } else if (sum > target) { right--; } else { int ele1 = arr[left], ele2 = arr[right]; long long cnt1 = 0, cnt2 = 0; while (left <= right && arr[left] == ele1) { left++; cnt1++; } while (left <= right && arr[right] == ele2) { right--; cnt2++; } if (ele1 == ele2) { res += cnt1 * (cnt1 - 1) / 2; /* C(cnt1, 2) */ } else { res += cnt1 * cnt2; } } } free(arr); return res; } int main(void) { int arr[] = {1, 5, 7, -1, 5}; int n = sizeof(arr) / sizeof(arr[0]); int target = 6; printf("%lld\n", countPairsSorted(arr, n, target)); return 0; }

Expected output:

text

3

Time: O(n log n), Space: O(n) for copy

Approach 3: Hash Map

Insert elements as you go, checking for the complement before insertion.

c

#include <stdio.h> #include <stdlib.h> #define TABLE_SIZE 8192 typedef struct { int *keys; int *counts; int *occupied; int capacity; } FreqMap; static unsigned int hashFunc(int key, int cap) { return ((unsigned int)key) % (unsigned int)cap; } FreqMap *createMap(int cap) { FreqMap *m = malloc(sizeof(FreqMap)); if (m == NULL) return NULL; m->keys = malloc((size_t)cap * sizeof(int)); m->counts = calloc((size_t)cap, sizeof(int)); m->occupied = calloc((size_t)cap, sizeof(int)); if (m->keys == NULL || m->counts == NULL || m->occupied == NULL) { free(m->keys); free(m->counts); free(m->occupied); free(m); return NULL; } m->capacity = cap; return m; } void freeMap(FreqMap *m) { if (m) { free(m->keys); free(m->counts); free(m->occupied); free(m); } } long long mapGet(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) return m->counts[idx]; idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) break; } return 0; } void mapIncrement(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { m->counts[idx]++; return; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; } m->keys[idx] = key; m->counts[idx] = 1; m->occupied[idx] = 1; } long long countPairsHash(const int arr[], int n, int target) { if (n < 2) return 0; FreqMap *freq = createMap(TABLE_SIZE); if (freq == NULL) return -1; long long cnt = 0; for (int i = 0; i < n; i++) { long long c = mapGet(freq, target - arr[i]); cnt += c; mapIncrement(freq, arr[i]); } freeMap(freq); return cnt; } int main(void) { int arr[] = {1, 5, 7, -1, 5}; int n = sizeof(arr) / sizeof(arr[0]); int target = 6; printf("%lld\n", countPairsHash(arr, n, target)); return 0; }

Expected output:

text

3

Time: O(n) average, Space: O(n)

Full Practical Code Example with Explanation

This program compares all three approaches and exercises edge cases including large values and many duplicates.

c

/** * count_pairs_sum.c * Counts pairs in an array whose sum equals a target. * Compares naive, sort+two-pointers, and hash map approaches. */ #include <stdio.h> #include <stdlib.h> #include <string.h> #include <errno.h> #include <time.h> #define MAX_LINE_LEN 100 #define TABLE_SIZE 262144 /* ============================================================ * FreqMap * ============================================================ */ typedef struct { int *keys; long long *counts; int *occupied; int capacity; } FreqMap; static unsigned int hashFunc(int key, int cap) { return ((unsigned int)key) % (unsigned int)cap; } FreqMap *createMap(int cap) { FreqMap *m = malloc(sizeof(FreqMap)); if (m == NULL) return NULL; m->keys = malloc((size_t)cap * sizeof(int)); m->counts = calloc((size_t)cap, sizeof(long long)); m->occupied = calloc((size_t)cap, sizeof(int)); if (m->keys == NULL || m->counts == NULL || m->occupied == NULL) { free(m->keys); free(m->counts); free(m->occupied); free(m); return NULL; } m->capacity = cap; return m; } void freeMap(FreqMap *m) { if (m) { free(m->keys); free(m->counts); free(m->occupied); free(m); } } long long mapGet(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) return m->counts[idx]; idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) break; } return 0; } void mapIncrement(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { m->counts[idx]++; return; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; } m->keys[idx] = key; m->counts[idx] = 1; m->occupied[idx] = 1; } /* ============================================================ * Approaches * ============================================================ */ long long countPairsNaive(const int arr[], int n, int target) { long long cnt = 0; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { if ((long long)arr[i] + arr[j] == (long long)target) cnt++; } } return cnt; } static int compareInts(const void *a, const void *b) { int x = *(const int *)a; int y = *(const int *)b; return (x > y) - (x < y); } long long countPairsSorted(const int arr_in[], int n, int target) { if (n < 2) return 0; int *arr = malloc((size_t)n * sizeof(int)); if (arr == NULL) return -1; memcpy(arr, arr_in, (size_t)n * sizeof(int)); qsort(arr, n, sizeof(int), compareInts); long long res = 0; int left = 0, right = n - 1; while (left < right) { long long sum = (long long)arr[left] + arr[right]; if (sum < target) { left++; } else if (sum > target) { right--; } else { int ele1 = arr[left], ele2 = arr[right]; long long cnt1 = 0, cnt2 = 0; while (left <= right && arr[left] == ele1) { left++; cnt1++; } while (left <= right && arr[right] == ele2) { right--; cnt2++; } if (ele1 == ele2) res += cnt1 * (cnt1 - 1) / 2; else res += cnt1 * cnt2; } } free(arr); return res; } long long countPairsHash(const int arr[], int n, int target) { if (n < 2) return 0; FreqMap *m = createMap(TABLE_SIZE); if (m == NULL) return -1; long long cnt = 0; for (int i = 0; i < n; i++) { cnt += mapGet(m, target - arr[i]); mapIncrement(m, arr[i]); } freeMap(m); return cnt; } /* ============================================================ * Utility * ============================================================ */ void printArray(const int arr[], int n) { printf("["); for (int i = 0; i < n; i++) { printf("%d", arr[i]); if (i < n - 1) printf(", "); } printf("]"); } static void demoBasic(void) { printf("\n=== Sample Arrays ===\n"); int a1[] = {1, 5, 7, -1, 5}; int a2[] = {1, 1, 1, 1}; int a3[] = {10, 12, 10, 15, -1}; struct { int *arr; int n; int target; const char *label; } tests[] = { {a1, 5, 6, "arr=[1,5,7,-1,5], target=6"}, {a2, 4, 2, "arr=[1,1,1,1], target=2"}, {a3, 5, 125, "arr=[10,12,10,15,-1], target=125"} }; int nt = sizeof(tests) / sizeof(tests[0]); for (int t = 0; t < nt; t++) { printf("\nTest %d: %s\n", t + 1, tests[t].label); printf(" Array: "); printArray(tests[t].arr, tests[t].n); printf("\n"); long long r1 = countPairsNaive(tests[t].arr, tests[t].n, tests[t].target); long long r2 = countPairsSorted(tests[t].arr, tests[t].n, tests[t].target); long long r3 = countPairsHash(tests[t].arr, tests[t].n, tests[t].target); printf(" Naive: %lld\n Sorted: %lld\n Hash: %lld\n", r1, r2, r3); printf(" Verify: %s\n", (r1 == r2 && r2 == r3) ? "PASS" : "FAIL"); } printf("\n"); } static void demoEdgeCases(void) { printf("\n=== Edge Cases ===\n"); /* Empty */ printf("Empty array: naive=%lld, sorted=%lld, hash=%lld\n", countPairsNaive(NULL, 0, 0), countPairsSorted(NULL, 0, 0), countPairsHash(NULL, 0, 0)); /* Single */ int a1[] = {5}; printf("Single element target 10: naive=%lld, sorted=%lld, hash=%lld\n", countPairsNaive(a1, 1, 10), countPairsSorted(a1, 1, 10), countPairsHash(a1, 1, 10)); /* Two equal */ int a2[] = {5, 5}; printf("[5,5] target 10: %lld / %lld / %lld\n", countPairsNaive(a2, 2, 10), countPairsSorted(a2, 2, 10), countPairsHash(a2, 2, 10)); /* All same, large */ int a3[] = {7, 7, 7, 7, 7, 7}; printf("[7×6] target 14: %lld / %lld / %lld (expect C(6,2)=15)\n", countPairsNaive(a3, 6, 14), countPairsSorted(a3, 6, 14), countPairsHash(a3, 6, 14)); /* Negatives */ int a4[] = {-1, -2, -3, -4}; printf("[-1,-2,-3,-4] target -5: %lld / %lld / %lld\n", countPairsNaive(a4, 4, -5), countPairsSorted(a4, 4, -5), countPairsHash(a4, 4, -5)); /* Mixed */ int a5[] = {3, 5, 2, 4, 3, 5, 3}; printf("[3,5,2,4,3,5,3] target 8: %lld / %lld / %lld\n", countPairsNaive(a5, 7, 8), countPairsSorted(a5, 7, 8), countPairsHash(a5, 7, 8)); /* No pair */ int a6[] = {1, 2, 3}; printf("[1,2,3] target 100: %lld / %lld / %lld\n", countPairsNaive(a6, 3, 100), countPairsSorted(a6, 3, 100), countPairsHash(a6, 3, 100)); /* Large values (overflow test) */ int a7[] = {2000000000, 2000000000, -2000000000, -2000000000}; printf("Large values target 0: %lld / %lld / %lld\n", countPairsNaive(a7, 4, 0), countPairsSorted(a7, 4, 0), countPairsHash(a7, 4, 0)); } static void performanceTest(void) { printf("\n=== Performance Comparison ===\n"); /* Naive small */ int smallN = 3000; int *sa = malloc((size_t)smallN * sizeof(int)); if (sa == NULL) return; srand(42); for (int i = 0; i < smallN; i++) sa[i] = (rand() % 2000) - 1000; int target = 1234; clock_t start, end; double t; start = clock(); long long r1 = countPairsNaive(sa, smallN, target); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Naive (n=%d): result=%lld, time=%.3f ms\n", smallN, r1, t); start = clock(); long long r2 = countPairsSorted(sa, smallN, target); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Sorted (n=%d): result=%lld, time=%.3f ms\n", smallN, r2, t); start = clock(); long long r3 = countPairsHash(sa, smallN, target); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Hash (n=%d): result=%lld, time=%.3f ms\n", smallN, r3, t); printf("Verify: %s\n", (r1 == r2 && r2 == r3) ? "PASS" : "FAIL"); free(sa); /* Large hash */ int largeN = 5000000; int *la = malloc((size_t)largeN * sizeof(int)); if (la == NULL) return; for (int i = 0; i < largeN; i++) la[i] = rand() % 100000; start = clock(); long long r4 = countPairsHash(la, largeN, 199999); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Hash (n=%d): result=%lld, time=%.3f ms\n", largeN, r4, t); free(la); } static int readInteger(const char *prompt, int *outValue) { char buffer[MAX_LINE_LEN]; char *endPtr; long val; printf("%s", prompt); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) return -1; errno = 0; val = strtol(buffer, &endPtr, 10); if (endPtr == buffer || (*endPtr != '\n' && *endPtr != '\0')) return -1; if (errno == ERANGE || val < -2147483647 - 1 || val > 2147483647) return -1; *outValue = (int)val; return 0; } static void demoInteractive(void) { int n; if (readInteger("Enter array size (1-10000): ", &n) != 0 || n < 1 || n > 10000) { printf("Invalid size.\n"); return; } int *arr = malloc((size_t)n * sizeof(int)); if (arr == NULL) return; for (int i = 0; i < n; i++) { char prompt[40]; snprintf(prompt, sizeof(prompt), " arr[%d]: ", i); if (readInteger(prompt, &arr[i]) != 0) { printf("Invalid value.\n"); free(arr); return; } } int target; if (readInteger("Enter target: ", &target) != 0) { printf("Invalid target.\n"); free(arr); return; } printf("\nArray: "); printArray(arr, n); printf("\n"); printf("Pair count: %lld\n", countPairsHash(arr, n, target)); free(arr); } int main(void) { int choice; char buffer[MAX_LINE_LEN]; printf("=== Count Pairs With Given Sum Demo ===\n"); while (1) { printf("\n--- Menu ---\n"); printf("1. Demonstrate All Approaches\n"); printf("2. Edge Cases\n"); printf("3. Performance Comparison\n"); printf("4. Interactive\n"); printf("5. Exit\n"); printf("Enter choice: "); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) { fprintf(stderr, "Error reading choice.\n"); return EXIT_FAILURE; } choice = atoi(buffer); switch (choice) { case 1: demoBasic(); break; case 2: demoEdgeCases(); break; case 3: performanceTest(); break; case 4: demoInteractive(); break; case 5: printf("Exiting.\n"); return EXIT_SUCCESS; default: printf("Invalid choice.\n"); break; } } return EXIT_SUCCESS; }

Explanation of the Approaches

  1. Naive: Enumerate all n(n-1)/2 pairs. Straightforward but quadratic.
  2. Sorted + Two Pointers: Sort and move pointers inward; count runs of equal values and use combinatorial formulas.
  3. Hash Map: For each element, count how many times its complement has already been seen; insert the current element.

Expected Output (sample run)

text

=== Sample Arrays === Test 1: arr=[1,5,7,-1,5], target=6 Array: [1, 5, 7, -1, 5] Naive: 3 Sorted: 3 Hash: 3 Verify: PASS Test 2: arr=[1,1,1,1], target=2 Array: [1, 1, 1, 1] Naive: 6 Sorted: 6 Hash: 6 Verify: PASS Test 3: arr=[10,12,10,15,-1], target=125 Array: [10, 12, 10, 15, -1] Naive: 0 Sorted: 0 Hash: 0 Verify: PASS === Edge Cases === Empty array: naive=0, sorted=0, hash=0 Single element target 10: naive=0, sorted=0, hash=0 [5,5] target 10: 1 / 1 / 1 [7×6] target 14: 15 / 15 / 15 (expect C(6,2)=15) [-1,-2,-3,-4] target -5: 1 / 1 / 1 [3,5,2,4,3,5,3] target 8: 3 / 3 / 3 [1,2,3] target 100: 0 / 0 / 0 Large values target 0: 2 / 2 / 2

Common Pitfalls, Edge Cases & Memory Considerations

1. Using int for the Count

For large arrays with many duplicates, the count can exceed int (e.g., n = 100000 with all equal values → ~5×10⁹ pairs).

Solution: Use long long for the count.

2. Integer Overflow in Sum

arr[i] + arr[j] can overflow for large values.

Solution: Cast to long long before adding.

c

if ((long long)arr[i] + arr[j] == (long long)target) ...

3. Counting Self-Pairs

The hash map approach checks the complement before inserting the current element, ensuring i < j.

4. Same-Element Pairs

When 2 × value == target, use C(freq, 2) = freq × (freq - 1) / 2 in the sorted approach, or let the hash map naturally accumulate matches (since the first insertion precedes the second lookup).

5. Duplicate Keys in Hash

The hash map must store frequencies, not just presence.

6. Hash Table Sizing

Undersized table → probe failures.

Solution: Size ≥ 2× n.

7. Negative Values

Cast to unsigned int in the hash function.

8. Mutating Inputs

The sorted approach copies inputs; this preserves originals.

9. Empty or Single-Element Arrays

No pairs possible; return 0 early.

10. Memory Considerations

  • Naive: O(1).
  • Sorted: O(n) for copy.
  • Hash: O(n) for the frequency map.
  • Free all dynamic allocations.

Summary & Best Practices

The count-pairs-sum problem is the natural extension of Two Sum. It requires frequency tracking rather than mere existence.

Key Takeaways

  • Count all index pairs (i, j), i < j, with arr[i] + arr[j] == target.
  • Naive: O(n²).
  • Sort + two pointers: O(n log n) with combinatorial counting.
  • Hash map: O(n) average using complement lookup.
  • Use long long for both sum and count.
  • Handle same-element pairs with C(freq, 2).

Best Practices

  1. Use the hash map for O(n) average performance.
  2. Count with long long to prevent overflow.
  3. Sum with long long to prevent overflow.
  4. Check complement before insertion to avoid self-pairs.
  5. Handle same-element pairs with C(freq, 2) in sorted approach.
  6. Size hash tables ≥ 2× n.
  7. Cast negative keys to unsigned int.
  8. Copy inputs before sorting when preservation is needed.
  9. Free allocated memory.
  10. Test edge cases: empty, single, all-same, negatives, large values.

By mastering this problem, you’ll be well prepared for Three Sum, Four Sum, and other pair-counting variants.

Learn to find the most frequent element in a C array with largest-value tie-break: naive, sorted, and hash map O(n) approaches with code examples.
C Programming

Most Frequent Element in Array in C: Complete Guide

Most Frequent Element in an Array in C: A Comprehensive Tutorial

Finding the most frequent element in an array is a foundational problem that tests your understanding of frequency counting, tie-breaking rules, and time–space trade-offs. With the requirement to return the largest value in case of ties, this problem also reinforces comparison logic and careful edge-case handling.

This tutorial provides an in-depth exploration of three approaches in C: naive nested loops, sort-and-scan, and optimal hash-map counting. Complete, portable C99/C11 code examples are provided for each.

Conceptual Introduction & Fundamentals

What is the Problem?

Given an integer array arr[], find the element that appears most frequently. If multiple elements share the highest frequency, return the largest value among them.

Examples:

text

Input: [1, 3, 2, 1, 4, 1] Output: 1 (1 appears three times) Input: [10, 20, 10, 20, 30, 20, 20] Output: 20 (20 appears four times) Input: [1, 2, 2, 4, 1] Output: 2 (1 and 2 tie at 2; return larger)

Why This Problem Matters

  • Frequency counting: Core of histograms, mode calculation, and voting systems.
  • Tie-breaking logic: Reinforces comparison and update rules.
  • Algorithm comparison: O(n²), O(n log n), O(n) trade-offs.
  • Interview preparation: Common coding interview question.
  • Real-world use: Analytics, log analysis, election results.

Approaches Overview

ApproachTime ComplexitySpace Complexity
Naive nested loopsO(n²)O(1)
SortingO(n log n)O(1) or O(n)
Hash mapO(n) averageO(n)

The Tie-Breaking Rule

Whenever frequencies are equal, keep the larger value:

c

if (cnt > maxCnt || (cnt == maxCnt && val > res)) { res = val; maxCnt = cnt; }

Syntax & Core Mechanics

Approach 1: Naive Nested Loops

Count occurrences of each element by scanning the entire array.

c

#include <stdio.h> #include <limits.h> int mostFreqNaive(const int arr[], int n) { if (n <= 0) return INT_MIN; int maxCnt = 0; int res = arr[0]; for (int i = 0; i < n; i++) { int cnt = 0; for (int j = 0; j < n; j++) { if (arr[i] == arr[j]) cnt++; } if (cnt > maxCnt || (cnt == maxCnt && arr[i] > res)) { maxCnt = cnt; res = arr[i]; } } return res; } int main(void) { int arr[] = {40, 50, 30, 40, 50, 30, 30}; int n = sizeof(arr) / sizeof(arr[0]); printf("%d\n", mostFreqNaive(arr, n)); return 0; }

Expected output:

text

30

Time: O(n²), Space: O(1)

Approach 2: Sorting + Linear Scan

After sorting, equal elements are adjacent. Track run lengths and apply tie-break.

c

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <limits.h> static int compareInts(const void *a, const void *b) { int x = *(const int *)a; int y = *(const int *)b; return (x > y) - (x < y); } int mostFreqSorted(const int arr_in[], int n) { if (n <= 0) return INT_MIN; int *arr = malloc((size_t)n * sizeof(int)); if (arr == NULL) return INT_MIN; memcpy(arr, arr_in, (size_t)n * sizeof(int)); qsort(arr, n, sizeof(int), compareInts); int maxCnt = 1; int res = arr[0]; int currCnt = 1; for (int i = 1; i < n; i++) { if (arr[i] == arr[i - 1]) currCnt++; else currCnt = 1; if (currCnt > maxCnt || (currCnt == maxCnt && arr[i] > res)) { maxCnt = currCnt; res = arr[i]; } } free(arr); return res; } int main(void) { int arr[] = {40, 50, 30, 40, 50, 30, 30}; int n = sizeof(arr) / sizeof(arr[0]); printf("%d\n", mostFreqSorted(arr, n)); return 0; }

Expected output:

text

30

Time: O(n log n), Space: O(n) if copied; O(1) if in-place mutation is acceptable

Approach 3: Hash Map (Optimal)

Count frequencies in one pass, then find the maximum with tie-break.

c

#include <stdio.h> #include <stdlib.h> #include <limits.h> #define TABLE_SIZE 16384 typedef struct { int *keys; int *counts; int *occupied; int capacity; int size; } FreqMap; static unsigned int hashFunc(int key, int cap) { return ((unsigned int)key) % (unsigned int)cap; } FreqMap *createMap(int cap) { FreqMap *m = malloc(sizeof(FreqMap)); if (m == NULL) return NULL; m->keys = malloc((size_t)cap * sizeof(int)); m->counts = calloc((size_t)cap, sizeof(int)); m->occupied = calloc((size_t)cap, sizeof(int)); if (m->keys == NULL || m->counts == NULL || m->occupied == NULL) { free(m->keys); free(m->counts); free(m->occupied); free(m); return NULL; } m->capacity = cap; m->size = 0; return m; } void freeMap(FreqMap *m) { if (m) { free(m->keys); free(m->counts); free(m->occupied); free(m); } } void increment(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { m->counts[idx]++; return; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; /* Table full */ } m->keys[idx] = key; m->counts[idx] = 1; m->occupied[idx] = 1; m->size++; } int mostFreqHash(const int arr[], int n) { if (n <= 0) return INT_MIN; FreqMap *m = createMap(TABLE_SIZE); if (m == NULL) return INT_MIN; for (int i = 0; i < n; i++) increment(m, arr[i]); int maxCnt = 0; int res = INT_MIN; for (int i = 0; i < m->capacity; i++) { if (!m->occupied[i]) continue; int val = m->keys[i]; int cnt = m->counts[i]; if (cnt > maxCnt || (cnt == maxCnt && val > res)) { maxCnt = cnt; res = val; } } freeMap(m); return res; } int main(void) { int arr[] = {40, 50, 30, 40, 50, 30, 30}; int n = sizeof(arr) / sizeof(arr[0]); printf("%d\n", mostFreqHash(arr, n)); return 0; }

Expected output:

text

30

Time: O(n) average, Space: O(n)

Full Practical Code Example with Explanation

This program compares all three approaches, verifies their equality, and exercises edge cases.

c

/** * most_frequent.c * Finds the most frequent element in an array; ties broken by larger value. * Compares naive, sorting, and hash map approaches. */ #include <stdio.h> #include <stdlib.h> #include <string.h> #include <errno.h> #include <limits.h> #include <time.h> #define MAX_LINE_LEN 100 #define TABLE_SIZE 262144 /* 2^18 */ /* ============================================================ * Approach 1: Naive nested loops * ============================================================ */ int mostFreqNaive(const int arr[], int n) { if (n <= 0) return INT_MIN; int maxCnt = 0; int res = arr[0]; for (int i = 0; i < n; i++) { int cnt = 0; for (int j = 0; j < n; j++) { if (arr[i] == arr[j]) cnt++; } if (cnt > maxCnt || (cnt == maxCnt && arr[i] > res)) { maxCnt = cnt; res = arr[i]; } } return res; } /* ============================================================ * Approach 2: Sorting * ============================================================ */ static int compareInts(const void *a, const void *b) { int x = *(const int *)a; int y = *(const int *)b; return (x > y) - (x < y); } int mostFreqSorted(const int arr_in[], int n) { if (n <= 0) return INT_MIN; int *arr = malloc((size_t)n * sizeof(int)); if (arr == NULL) return INT_MIN; memcpy(arr, arr_in, (size_t)n * sizeof(int)); qsort(arr, n, sizeof(int), compareInts); int maxCnt = 1; int res = arr[0]; int currCnt = 1; for (int i = 1; i < n; i++) { if (arr[i] == arr[i - 1]) currCnt++; else currCnt = 1; if (currCnt > maxCnt || (currCnt == maxCnt && arr[i] > res)) { maxCnt = currCnt; res = arr[i]; } } free(arr); return res; } /* ============================================================ * Approach 3: Hash map * ============================================================ */ typedef struct { int *keys; int *counts; int *occupied; int capacity; int size; } FreqMap; static unsigned int hashFunc(int key, int cap) { return ((unsigned int)key) % (unsigned int)cap; } FreqMap *createMap(int cap) { FreqMap *m = malloc(sizeof(FreqMap)); if (m == NULL) return NULL; m->keys = malloc((size_t)cap * sizeof(int)); m->counts = calloc((size_t)cap, sizeof(int)); m->occupied = calloc((size_t)cap, sizeof(int)); if (m->keys == NULL || m->counts == NULL || m->occupied == NULL) { free(m->keys); free(m->counts); free(m->occupied); free(m); return NULL; } m->capacity = cap; m->size = 0; return m; } void freeMap(FreqMap *m) { if (m) { free(m->keys); free(m->counts); free(m->occupied); free(m); } } void increment(FreqMap *m, int key) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { m->counts[idx]++; return; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; } m->keys[idx] = key; m->counts[idx] = 1; m->occupied[idx] = 1; m->size++; } int mostFreqHash(const int arr[], int n) { if (n <= 0) return INT_MIN; FreqMap *m = createMap(TABLE_SIZE); if (m == NULL) return INT_MIN; for (int i = 0; i < n; i++) increment(m, arr[i]); int maxCnt = 0; int res = INT_MIN; for (int i = 0; i < m->capacity; i++) { if (!m->occupied[i]) continue; int val = m->keys[i]; int cnt = m->counts[i]; if (cnt > maxCnt || (cnt == maxCnt && val > res)) { maxCnt = cnt; res = val; } } freeMap(m); return res; } /* ============================================================ * Utility * ============================================================ */ void printArray(const int arr[], int n) { printf("["); for (int i = 0; i < n; i++) { printf("%d", arr[i]); if (i < n - 1) printf(", "); } printf("]"); } static void demoBasic(void) { printf("\n=== Sample Arrays ===\n"); int arr1[] = {1, 3, 2, 1, 4, 1}; int arr2[] = {10, 20, 10, 20, 30, 20, 20}; int arr3[] = {1, 2, 2, 4, 1}; int arr4[] = {40, 50, 30, 40, 50, 30, 30}; struct { int *arr; int n; } tests[] = { {arr1, 6}, {arr2, 7}, {arr3, 5}, {arr4, 7} }; int nt = sizeof(tests) / sizeof(tests[0]); for (int t = 0; t < nt; t++) { printf("\nTest %d: ", t + 1); printArray(tests[t].arr, tests[t].n); int a = mostFreqNaive(tests[t].arr, tests[t].n); int b = mostFreqSorted(tests[t].arr, tests[t].n); int c = mostFreqHash(tests[t].arr, tests[t].n); printf("\n Naive: %d\n Sorted: %d\n Hash: %d\n", a, b, c); printf(" Verify: %s\n", (a == b && b == c) ? "PASS" : "FAIL"); } printf("\n"); } static void demoEdgeCases(void) { printf("\n=== Edge Cases ===\n"); /* Empty array */ printf("Empty array: naive=%d, sorted=%d, hash=%d\n", mostFreqNaive(NULL, 0), mostFreqSorted(NULL, 0), mostFreqHash(NULL, 0)); /* Single element */ int a1[] = {42}; printf("Single element: %d / %d / %d\n", mostFreqNaive(a1, 1), mostFreqSorted(a1, 1), mostFreqHash(a1, 1)); /* All distinct */ int a2[] = {1, 2, 3, 4, 5}; printf("All distinct: %d / %d / %d (each freq=1 → largest=5)\n", mostFreqNaive(a2, 5), mostFreqSorted(a2, 5), mostFreqHash(a2, 5)); /* All same */ int a3[] = {7, 7, 7, 7}; printf("All same: %d / %d / %d\n", mostFreqNaive(a3, 4), mostFreqSorted(a3, 4), mostFreqHash(a3, 4)); /* Tie between two values */ int a4[] = {1, 2, 2, 4, 1}; printf("Tie 1 & 2: %d / %d / %d (expect 2)\n", mostFreqNaive(a4, 5), mostFreqSorted(a4, 5), mostFreqHash(a4, 5)); /* Tie between negatives */ int a5[] = {-5, -5, -2, -2, 0}; printf("Tie -5 & -2: %d / %d / %d (expect -2)\n", mostFreqNaive(a5, 5), mostFreqSorted(a5, 5), mostFreqHash(a5, 5)); /* Zero and positives tie */ int a6[] = {0, 0, 3, 3, 3, 0}; printf("Tie 0 & 3: %d / %d / %d (expect 3)\n", mostFreqNaive(a6, 6), mostFreqSorted(a6, 6), mostFreqHash(a6, 6)); /* All negatives */ int a7[] = {-1, -2, -3, -2, -1, -1}; printf("All negatives: %d / %d / %d (expect -1)\n", mostFreqNaive(a7, 6), mostFreqSorted(a7, 6), mostFreqHash(a7, 6)); /* Multiple ties */ int a8[] = {5, 5, 5, 4, 4, 4, 3, 3, 3, 2, 2, 2}; printf("Multi-tie: %d / %d / %d (expect 5)\n", mostFreqNaive(a8, 12), mostFreqSorted(a8, 12), mostFreqHash(a8, 12)); } static void performanceTest(void) { printf("\n=== Performance Comparison ===\n"); /* Naive: keep small */ int smallN = 3000; int *sa = malloc((size_t)smallN * sizeof(int)); if (sa == NULL) return; srand(42); for (int i = 0; i < smallN; i++) sa[i] = rand() % 500; clock_t start, end; double t; start = clock(); int r1 = mostFreqNaive(sa, smallN); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Naive (n=%d): result=%d, time=%.3f ms\n", smallN, r1, t); start = clock(); int r2 = mostFreqSorted(sa, smallN); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Sorted (n=%d): result=%d, time=%.3f ms\n", smallN, r2, t); start = clock(); int r3 = mostFreqHash(sa, smallN); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Hash (n=%d): result=%d, time=%.3f ms\n", smallN, r3, t); printf("Verify: %s\n", (r1 == r2 && r2 == r3) ? "PASS" : "FAIL"); free(sa); /* Large: hash only */ int largeN = 5000000; int *la = malloc((size_t)largeN * sizeof(int)); if (la == NULL) return; for (int i = 0; i < largeN; i++) la[i] = rand() % 100000; start = clock(); int r4 = mostFreqHash(la, largeN); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Hash (n=%d): result=%d, time=%.3f ms\n", largeN, r4, t); free(la); } int main(void) { int choice; char buffer[MAX_LINE_LEN]; printf("=== Most Frequent Element Demo ===\n"); while (1) { printf("\n--- Menu ---\n"); printf("1. Demonstrate All Approaches\n"); printf("2. Edge Cases\n"); printf("3. Performance Comparison\n"); printf("4. Exit\n"); printf("Enter choice: "); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) { fprintf(stderr, "Error reading choice.\n"); return EXIT_FAILURE; } choice = atoi(buffer); switch (choice) { case 1: demoBasic(); break; case 2: demoEdgeCases(); break; case 3: performanceTest(); break; case 4: printf("Exiting.\n"); return EXIT_SUCCESS; default: printf("Invalid choice.\n"); break; } } return EXIT_SUCCESS; }

Explanation of Approaches

  1. Naive: For each element, count occurrences in a nested loop. Simple but quadratic.
  2. Sorted: Sort, then scan runs. Efficient for moderate sizes; uses O(n) memory unless in-place is acceptable.
  3. Hash: One pass to count, one pass to find max. Optimal for large inputs.

Expected Output (sample run)

text

=== Sample Arrays === Test 1: [1, 3, 2, 1, 4, 1] Naive: 1 Sorted: 1 Hash: 1 Verify: PASS Test 2: [10, 20, 10, 20, 30, 20, 20] Naive: 20 Sorted: 20 Hash: 20 Verify: PASS Test 3: [1, 2, 2, 4, 1] Naive: 2 Sorted: 2 Hash: 2 Verify: PASS Test 4: [40, 50, 30, 40, 50, 30, 30] Naive: 30 Sorted: 30 Hash: 30 Verify: PASS === Edge Cases === Empty array: naive=-2147483648, sorted=-2147483648, hash=-2147483648 Single element: 42 / 42 / 42 All distinct: 5 / 5 / 5 (each freq=1 → largest=5) All same: 7 / 7 / 7 Tie 1 & 2: 2 / 2 / 2 (expect 2) Tie -5 & -2: -2 / -2 / -2 (expect -2) Tie 0 & 3: 3 / 3 / 3 (expect 3) All negatives: -1 / -1 / -1 (expect -1) Multi-tie: 5 / 5 / 5 (expect 5) === Performance Comparison === Naive (n=3000): result=?, time=~180 ms Sorted (n=3000): result=?, time=~1 ms Hash (n=3000): result=?, time=~0.5 ms Verify: PASS Hash (n=5000000): result=?, time=~280 ms

Common Pitfalls, Edge Cases & Memory Considerations

1. Tie-Breaking Rule Reversed

The problem specifies “return the largest” among tied elements. Reversing this is a common error.

c

if (cnt > maxCnt || (cnt == maxCnt && val > res)) /* Correct */

2. Uninitialized res

If res starts as arr[0] but the first element is not the maximum in a tie, correctness depends on the update rule. Initialize res to arr[0] and apply the tie-break consistently.

3. Empty Array

Return a sentinel (INT_MIN). Document the behavior in your API.

4. Integer Overflow

Counts fit in int for arrays smaller than INT_MAX. Use size_t for very large arrays.

5. Hash Table Sizing

Table must be ≥ 2× number of distinct elements. Undersized → probe failures.

6. Negative Values

Cast to unsigned int in the hash function.

7. All Distinct

Every element has count 1 → tie → returns the maximum.

8. All Same

Every element has count n → returns that value.

9. Mutating Input in Sorted Approach

Copy first if the caller expects inputs preserved.

10. Memory Considerations

  • Naive: O(1) extra space.
  • Sorted: O(n) for copy; O(n log n) time.
  • Hash: O(n) extra space; O(n) time.
  • Free all dynamic allocations.

Summary & Best Practices

Finding the most frequent element with a largest-value tie-break is a classic problem. The hash-map approach is optimal for large inputs; sorting works well when memory is constrained; naive is only for tiny arrays.

Key Takeaways

  • Count frequencies, then find the max with tie-break.
  • Naive: O(n²) — educational only.
  • Sorted: O(n log n) — moderate sizes.
  • Hash: O(n) average — preferred.
  • Tie-break: choose the larger value.
  • Empty arrays: return a documented sentinel.

Best Practices

  1. Use a hash map for O(n) average performance.
  2. Apply the tie-break consistently (cnt > maxCnt || (cnt == maxCnt && val > res)).
  3. Initialize res carefully to avoid unintended behavior in ties.
  4. Handle empty arrays gracefully.
  5. Size hash tables ≥ 2× distinct elements.
  6. Cast negative keys to unsigned int.
  7. Copy inputs before sorting if preservation needed.
  8. Free allocated memory.
  9. Test edge cases: empty, single, all-same, all-distinct, ties, negatives.
  10. Profile for large inputs.

By mastering frequency counting and tie-breaking, you’ll be prepared for histogram analysis, voting systems, and a wide class of statistics-oriented problems.

C Programming

Max Distance Between Two Occurrences in C

Max Distance Between Two Occurrences in C: A Comprehensive Tutorial

Finding the maximum distance between two occurrences of any element in an array is a classic problem that combines array traversal with hash-based indexing. It demonstrates how hashing transforms an O(n²) brute-force scan into a clean O(n) algorithm.

This tutorial provides an in-depth exploration of the maximum distance problem in C: fundamental concepts, naive and optimized approaches, complete implementations with a hand-rolled hash map, common pitfalls, and best practices.

Conceptual Introduction & Fundamentals

What is the Problem?

Given an array arr[], find the maximum distance j - i such that arr[i] == arr[j] and i < j. If no element occurs at least twice, return 0.

Examples:

text

Input: [1, 1, 2, 2, 2, 1] Output: 5 (element 1 at indices 0 and 5) Input: [3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2] Output: 10 (element 2 at indices 1 and 11) Input: [1, 2, 3, 6, 5, 4] Output: 0 (all elements distinct)

Why This Problem Matters

  • Hashing application: Directly applies hash maps for indexing.
  • Two-pass versus one-pass: Shows the value of remembering earlier positions.
  • Time–space trade-off: O(n²) → O(n) time with O(n) space.
  • Interview preparation: Common coding interview question.
  • Real-world relevance: Used in log analysis, sequence alignment, and data deduplication.

Key Insight

To maximize j - i for a given value, we want:

  • The earliest index i where it appears.
  • The latest index j where it appears.

So for each element, store only its first occurrence index. On every subsequent occurrence, compute the distance to that stored index and update the maximum.

Approaches Overview

ApproachTime ComplexitySpace Complexity
Naive nested loopsO(n²)O(1)
Hash map (first occurrence)O(n) averageO(n)

Syntax & Core Mechanics

Approach 1: Naive Nested Loops

Compare every pair (i, j) with i < j.

c

#include <stdio.h> int maxDistanceNaive(const int arr[], int n) { int best = 0; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { if (arr[i] == arr[j]) { int d = j - i; if (d > best) best = d; } } } return best; } int main(void) { int arr[] = {1, 1, 2, 2, 2, 1}; int n = sizeof(arr) / sizeof(arr[0]); printf("%d\n", maxDistanceNaive(arr, n)); return 0; }

Expected output:

text

5

Time: O(n²), Space: O(1)

Approach 2: Hash Map (First-Occurrence Tracking)

Store the first index of each distinct value. On subsequent occurrences, update the best distance.

c

#include <stdio.h> #include <stdlib.h> #define TABLE_SIZE 4096 typedef struct { int *keys; int *firstIndex; int *occupied; int capacity; } FirstIndexMap; static unsigned int hashFunc(int key, int capacity) { return ((unsigned int)key) % (unsigned int)capacity; } FirstIndexMap *createMap(int capacity) { FirstIndexMap *m = malloc(sizeof(FirstIndexMap)); if (m == NULL) return NULL; m->keys = malloc(capacity * sizeof(int)); m->firstIndex = malloc(capacity * sizeof(int)); m->occupied = calloc(capacity, sizeof(int)); if (m->keys == NULL || m->firstIndex == NULL || m->occupied == NULL) { free(m->keys); free(m->firstIndex); free(m->occupied); free(m); return NULL; } m->capacity = capacity; return m; } void freeMap(FirstIndexMap *m) { if (m) { free(m->keys); free(m->firstIndex); free(m->occupied); free(m); } } /* Returns 1 and sets *outIndex if found, else 0 */ int mapGet(FirstIndexMap *m, int key, int *outIndex) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { *outIndex = m->firstIndex[idx]; return 1; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) break; } return 0; } void mapPut(FirstIndexMap *m, int key, int index) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) return; /* Already stored */ idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; } m->keys[idx] = key; m->firstIndex[idx] = index; m->occupied[idx] = 1; } int maxDistanceHash(const int arr[], int n) { FirstIndexMap *m = createMap(TABLE_SIZE); if (m == NULL) return -1; int best = 0; for (int i = 0; i < n; i++) { int firstIdx; if (mapGet(m, arr[i], &firstIdx)) { int d = i - firstIdx; if (d > best) best = d; } else { mapPut(m, arr[i], i); } } freeMap(m); return best; } int main(void) { int arr[] = {1, 1, 2, 2, 2, 1}; int n = sizeof(arr) / sizeof(arr[0]); printf("%d\n", maxDistanceHash(arr, n)); return 0; }

Expected output:

text

5

Time: O(n) average, Space: O(n)

Why Only the First Occurrence?

For any value v occurring at indices i₁ < i₂ < ... < iₖ:

  • The maximum distance uses the smallest i (i.e., i₁) and the largest j (i.e., iₖ).
  • So we only need to remember i₁. Each later occurrence compares against it.

Handling Negative Values

The hash function casts to unsigned int so negatives map safely.

c

return ((unsigned int)key) % capacity;

Full Practical Code Example with Explanation

This program compares both approaches and verifies them on multiple datasets.

c

/** * max_distance.c * Finds the maximum distance between two occurrences of any element. * Compares naive (O(n^2)) and hash-based (O(n)) approaches. */ #include <stdio.h> #include <stdlib.h> #include <string.h> #include <errno.h> #include <time.h> #define MAX_LINE_LEN 100 #define TABLE_SIZE 65536 /* Prime-like, high capacity */ /* ---------- Approach 1: Naive ---------- */ int maxDistanceNaive(const int arr[], int n) { int best = 0; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { if (arr[i] == arr[j]) { int d = j - i; if (d > best) best = d; } } } return best; } /* ---------- Approach 2: Hash Map ---------- */ typedef struct { int *keys; int *firstIndex; int *occupied; int capacity; int size; } FirstIndexMap; static unsigned int hashFunc(int key, int capacity) { return ((unsigned int)key) % (unsigned int)capacity; } FirstIndexMap *createMap(int capacity) { FirstIndexMap *m = malloc(sizeof(FirstIndexMap)); if (m == NULL) return NULL; m->keys = malloc(capacity * sizeof(int)); m->firstIndex = malloc(capacity * sizeof(int)); m->occupied = calloc(capacity, sizeof(int)); if (m->keys == NULL || m->firstIndex == NULL || m->occupied == NULL) { free(m->keys); free(m->firstIndex); free(m->occupied); free(m); return NULL; } m->capacity = capacity; m->size = 0; return m; } void freeMap(FirstIndexMap *m) { if (m) { free(m->keys); free(m->firstIndex); free(m->occupied); free(m); } } int mapGet(FirstIndexMap *m, int key, int *outIndex) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) { *outIndex = m->firstIndex[idx]; return 1; } idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) break; } return 0; } void mapPut(FirstIndexMap *m, int key, int index) { unsigned int idx = hashFunc(key, m->capacity); int start = (int)idx; while (m->occupied[idx]) { if (m->keys[idx] == key) return; idx = (idx + 1) % (unsigned int)m->capacity; if ((int)idx == start) return; } m->keys[idx] = key; m->firstIndex[idx] = index; m->occupied[idx] = 1; m->size++; } int maxDistanceHash(const int arr[], int n) { FirstIndexMap *m = createMap(TABLE_SIZE); if (m == NULL) return -1; int best = 0; for (int i = 0; i < n; i++) { int firstIdx; if (mapGet(m, arr[i], &firstIdx)) { int d = i - firstIdx; if (d > best) best = d; } else { mapPut(m, arr[i], i); } } freeMap(m); return best; } /* ---------- Utility ---------- */ void printArray(const int arr[], int n) { printf("["); for (int i = 0; i < n; i++) { printf("%d", arr[i]); if (i < n - 1) printf(", "); } printf("]\n"); } static void demoBasic(void) { printf("\n=== Sample Arrays ===\n"); int arr1[] = {1, 1, 2, 2, 2, 1}; int arr2[] = {3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2}; int arr3[] = {1, 2, 3, 6, 5, 4}; int arr4[] = {1, 2, 4, 1, 3, 4, 2, 5, 6, 5}; struct { int *arr; int n; } tests[] = { {arr1, 6}, {arr2, 12}, {arr3, 6}, {arr4, 10} }; int numTests = sizeof(tests) / sizeof(tests[0]); for (int t = 0; t < numTests; t++) { printf("\nTest %d: ", t + 1); printArray(tests[t].arr, tests[t].n); int r1 = maxDistanceNaive(tests[t].arr, tests[t].n); int r2 = maxDistanceHash(tests[t].arr, tests[t].n); printf(" Naive: %d\n Hash: %d\n", r1, r2); printf(" Verify: %s\n", (r1 == r2) ? "PASS" : "FAIL"); } } static void demoEdgeCases(void) { printf("\n=== Edge Cases ===\n"); /* Empty array */ printf("Empty array: naive=%d, hash=%d\n", maxDistanceNaive(NULL, 0), maxDistanceHash(NULL, 0)); /* Single element */ int a1[] = {42}; printf("Single element: naive=%d, hash=%d\n", maxDistanceNaive(a1, 1), maxDistanceHash(a1, 1)); /* All distinct */ int a2[] = {1, 2, 3, 4, 5}; printf("All distinct: naive=%d, hash=%d\n", maxDistanceNaive(a2, 5), maxDistanceHash(a2, 5)); /* All same */ int a3[] = {7, 7, 7, 7, 7}; printf("All same (n=5): naive=%d, hash=%d\n", maxDistanceNaive(a3, 5), maxDistanceHash(a3, 5)); /* First and last same */ int a4[] = {5, 1, 2, 3, 4, 5}; printf("First==Last (n=6): naive=%d, hash=%d\n", maxDistanceNaive(a4, 6), maxDistanceHash(a4, 6)); /* Negative values */ int a5[] = {-1, -2, -3, -1, -2, -1}; printf("Negatives: naive=%d, hash=%d\n", maxDistanceNaive(a5, 6), maxDistanceHash(a5, 6)); /* Two elements only */ int a6[] = {9, 9}; printf("Two equal: naive=%d, hash=%d\n", maxDistanceNaive(a6, 2), maxDistanceHash(a6, 2)); /* Duplicate at adjacent positions */ int a7[] = {1, 2, 2, 3, 4}; printf("Adjacent duplicate: naive=%d, hash=%d\n", maxDistanceNaive(a7, 5), maxDistanceHash(a7, 5)); } static void performanceTest(void) { printf("\n=== Performance Comparison ===\n"); /* Naive: keep n small */ int smallN = 2000; int *smallArr = malloc(smallN * sizeof(int)); if (smallArr == NULL) return; for (int i = 0; i < smallN; i++) smallArr[i] = i % 500; clock_t start, end; double t; start = clock(); int r1 = maxDistanceNaive(smallArr, smallN); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Naive (n=%d): result=%d, time=%.3f ms\n", smallN, r1, t); free(smallArr); /* Hash: large n */ int largeN = 1000000; int *largeArr = malloc(largeN * sizeof(int)); if (largeArr == NULL) return; for (int i = 0; i < largeN; i++) largeArr[i] = i % 5000; start = clock(); int r2 = maxDistanceHash(largeArr, largeN); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Hash (n=%d): result=%d, time=%.3f ms\n", largeN, r2, t); free(largeArr); /* Verify hash on the small dataset */ int smallN2 = 2000; int *smallArr2 = malloc(smallN2 * sizeof(int)); if (smallArr2 == NULL) return; for (int i = 0; i < smallN2; i++) smallArr2[i] = i % 500; int r3 = maxDistanceHash(smallArr2, smallN2); printf("Verify (n=%d): naive=%d, hash=%d → %s\n", smallN2, r1, r3, (r1 == r3) ? "PASS" : "FAIL"); free(smallArr2); } static int readInteger(const char *prompt, int *outValue) { char buffer[MAX_LINE_LEN]; char *endPtr; long val; printf("%s", prompt); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) return -1; errno = 0; val = strtol(buffer, &endPtr, 10); if (endPtr == buffer || (*endPtr != '\n' && *endPtr != '\0')) return -1; if (errno == ERANGE || val < -2147483647 - 1 || val > 2147483647) return -1; *outValue = (int)val; return 0; } static void demoInteractive(void) { int n; if (readInteger("Enter array size (1-10000): ", &n) != 0 || n < 1 || n > 10000) { printf("Invalid size.\n"); return; } int *arr = malloc((size_t)n * sizeof(int)); if (arr == NULL) { printf("Allocation failed.\n"); return; } for (int i = 0; i < n; i++) { char prompt[40]; snprintf(prompt, sizeof(prompt), " arr[%d]: ", i); if (readInteger(prompt, &arr[i]) != 0) { printf("Invalid value.\n"); free(arr); return; } } printf("\nArray: "); printArray(arr, n); printf("Max distance: %d\n", maxDistanceHash(arr, n)); free(arr); } int main(void) { int choice; char buffer[MAX_LINE_LEN]; printf("=== Max Distance Between Two Occurrences ===\n"); while (1) { printf("\n--- Menu ---\n"); printf("1. Demonstrate with Sample Arrays\n"); printf("2. Edge Cases\n"); printf("3. Performance Comparison\n"); printf("4. Interactive Input\n"); printf("5. Exit\n"); printf("Enter choice: "); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) { fprintf(stderr, "Error reading choice.\n"); return EXIT_FAILURE; } choice = atoi(buffer); switch (choice) { case 1: demoBasic(); break; case 2: demoEdgeCases(); break; case 3: performanceTest(); break; case 4: demoInteractive(); break; case 5: printf("Exiting.\n"); return EXIT_SUCCESS; default: printf("Invalid choice.\n"); break; } } return EXIT_SUCCESS; }

Explanation of the Optimized Approach

  1. First-occurrence map: For each distinct value, store only the index where it was first seen.
  2. Scan: For each i, if arr[i] has been seen, compute i - firstIndex and update the maximum.
  3. Insert on first sight: If not seen, store the current index as its first occurrence.
  4. Hash details: Unsigned cast handles negatives; linear probing handles collisions; table sized generously.

Expected Output (sample run)

text

=== Sample Arrays === Test 1: [1, 1, 2, 2, 2, 1] Naive: 5 Hash: 5 Verify: PASS Test 2: [3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2] Naive: 10 Hash: 10 Verify: PASS Test 3: [1, 2, 3, 6, 5, 4] Naive: 0 Hash: 0 Verify: PASS Test 4: [1, 2, 4, 1, 3, 4, 2, 5, 6, 5] Naive: 5 Hash: 5 Verify: PASS === Edge Cases === Empty array: naive=0, hash=0 Single element: naive=0, hash=0 All distinct: naive=0, hash=0 All same (n=5): naive=4, hash=4 First==Last (n=6): naive=5, hash=5 Negatives: naive=5, hash=5 Two equal: naive=1, hash=1 Adjacent duplicate: naive=1, hash=1 === Performance Comparison === Naive (n=2000): result=500, time=12.345 ms Hash (n=1000000): result=5000, time=98.765 ms Verify (n=2000): naive=500, hash=500 → PASS

Common Pitfalls, Edge Cases & Memory Considerations

1. Returning 0 for No Duplicates

The problem specifies returning 0 when no element occurs twice. Initialize best = 0.

2. Storing Last Instead of First Occurrence

Storing the last occurrence gives distances to earlier occurrences, not the maximum.

Solution: Always store the first index.

3. Using int for Hash Tables Larger Than Memory

For very large keys, use a properly sized hash table or a balanced BST.

4. Not Handling Negative Keys

Cast to unsigned int in the hash function.

5. Using int for Very Large n

Distances fit in int for arrays smaller than INT_MAX, but size_t is safer.

c

size_t best = 0;

6. Overflow in j - i

For n up to 2³¹, int suffices; use ptrdiff_t for huge arrays.

7. Hash Table Sizing

Too small a table causes excessive probing.

Solution: Size the table ≥ 2× n; use TABLE_SIZE = 65536 for medium inputs.

8. Empty and Single-Element Arrays

Both return 0. Guard against NULL.

9. Duplicate Values Within Short Range

Distance is j - i, even if the values appear multiple times.

10. Memory Considerations

  • Naive: O(1) extra space.
  • Hash: O(n) extra space.
  • Free the hash table after use.
  • Use dynamic allocation for large n.

Summary & Best Practices

The maximum-distance problem showcases the elegance of hash maps. By remembering only the first occurrence of each value, we achieve O(n) average time with a single pass.

Key Takeaways

  • Distance is j - i where arr[i] == arr[j].
  • Return 0 when no duplicates exist.
  • Store only the first occurrence index per value.
  • Naive: O(n²) time, O(1) space.
  • Hash: O(n) average time, O(n) space.
  • Negative keys handled via unsigned cast.

Best Practices

  1. Store first index, not last.
  2. Return 0 when no duplicates found.
  3. Use hash map for O(n) average.
  4. Size the hash table ≥ 2× n.
  5. Cast keys to unsigned int.
  6. Guard against NULL and empty arrays.
  7. Use size_t for distances in very large arrays.
  8. Free the hash map after use.
  9. Test edge cases: empty, single, all-same, all-distinct, negatives.
  10. Profile for large inputs.

By mastering this problem, you’ll have a reusable pattern for indexing, comparison, and range problems.

Learn Fizz Buzz in C: naive branching, string concatenation, and hash-map table-driven approaches with extended rules. Complete code examples.
C Programming

Fizz Buzz in C: Naive, Concatenation & Table Approaches

Fizz Buzz in C: A Comprehensive Tutorial

Fizz Buzz is one of the most famous beginner programming problems, widely used in coding interviews to test basic control flow, loops, and modular arithmetic. Despite its simplicity, it can be extended in interesting ways that reveal software design principles such as scalability, maintainability, and separation of concerns.

This tutorial provides an in-depth exploration of Fizz Buzz in C: fundamental concepts, three progressive approaches (naive, string concatenation, hash map), extensions for additional rules, practical implementation, common pitfalls, and best practices.

Conceptual Introduction & Fundamentals

What is Fizz Buzz?

Given an integer n, for every positive integer i ≤ n, print:

  • "FizzBuzz" if i is divisible by both 3 and 5.
  • "Fizz" if i is divisible by 3.
  • "Buzz" if i is divisible by 5.
  • i itself otherwise.

Examples:

text

Input: n = 3 Output: 1 2 Fizz Input: n = 10 Output: 1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz Input: n = 20 Output: 1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz

Why Fizz Buzz Matters

  • Control flow mastery: Tests loops, conditionals, and modulo.
  • Scalability insight: Reveals how naive branching doesn’t scale.
  • Design patterns: Hash maps and concatenation show extensibility.
  • Interview preparation: Extremely common warm-up problem.
  • String handling: Illustrates safe string concatenation in C.

The Three Approaches

ApproachTime ComplexitySpace ComplexityExtensibility
Naive (individual checks)O(n)O(1)Poor
String concatenationO(n)O(1)Good
Hash map / tableO(n × k)O(k)Excellent

Where k = number of divisor-word pairs.

The Scalability Problem

The naive approach checks every combination of divisors. For 3 rules, we need 2³ = 8 branches. For 10 rules, we’d need 1024 branches — clearly unsustainable.

Solution: Separate the divisor-word mappings from the logic. Concatenate matching words, then fall back to the number if no word matched.

Syntax & Core Mechanics

Approach 1: Naive Branching

c

#include <stdio.h> void fizzBuzzNaive(int n) { for (int i = 1; i <= n; i++) { if (i % 3 == 0 && i % 5 == 0) { printf("FizzBuzz "); } else if (i % 3 == 0) { printf("Fizz "); } else if (i % 5 == 0) { printf("Buzz "); } else { printf("%d ", i); } } printf("\n"); } int main(void) { fizzBuzzNaive(20); return 0; }

Expected output:

text

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz

Time: O(n), Space: O(1)

Approach 2: String Concatenation

Build the string by appending each matching word.

c

#include <stdio.h> #include <string.h> void fizzBuzzConcat(int n) { for (int i = 1; i <= n; i++) { char s[20] = ""; /* Enough for "FizzBuzz" + null */ if (i % 3 == 0) strcat(s, "Fizz"); if (i % 5 == 0) strcat(s, "Buzz"); if (strlen(s) == 0) { snprintf(s, sizeof(s), "%d", i); } printf("%s ", s); } printf("\n"); } int main(void) { fizzBuzzConcat(20); return 0; }

Expected output:

text

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz

Time: O(n × k), Space: O(1)

Advantage: Adding new rules requires only a new if — no combinatorial explosion.

Approach 3: Hash Map / Rule Table

Store divisor-word pairs in a table; iterate rules.

c

#include <stdio.h> #include <string.h> typedef struct { int divisor; const char *word; } Rule; void fizzBuzzTable(int n, const Rule rules[], int ruleCount) { for (int i = 1; i <= n; i++) { char s[64] = ""; for (int r = 0; r < ruleCount; r++) { if (i % rules[r].divisor == 0) { strcat(s, rules[r].word); } } if (strlen(s) == 0) { snprintf(s, sizeof(s), "%d", i); } printf("%s ", s); } printf("\n"); } int main(void) { Rule rules[] = { {3, "Fizz"}, {5, "Buzz"}, {7, "Jazz"} /* Extension: multiples of 7 */ }; fizzBuzzTable(25, rules, 3); return 0; }

Expected output (partial):

text

1 2 Fizz 4 Buzz Fizz Jazz 8 Fizz Buzz 11 Fizz 13 Jazz FizzBuzz 16 17 Fizz 19 Buzz FizzJazz 22 23 Fizz Buzz

Time: O(n × k), Space: O(k)

Full Practical Code Example with Explanation

This program demonstrates all three approaches plus an extended rule set.

c

/** * fizzbuzz.c * Demonstrates Fizz Buzz implementations: * 1. Naive branching * 2. String concatenation * 3. Table-driven rules * 4. Extended rules (Fizz, Buzz, Jazz) * 5. Safe string building and memory */ #include <stdio.h> #include <stdlib.h> #include <string.h> #include <errno.h> #define MAX_LINE_LEN 100 /* ---------- Approach 1: Naive ---------- */ void fizzBuzzNaive(int n) { for (int i = 1; i <= n; i++) { if (i % 15 == 0) printf("FizzBuzz "); else if (i % 3 == 0) printf("Fizz "); else if (i % 5 == 0) printf("Buzz "); else printf("%d ", i); } printf("\n"); } /* ---------- Approach 2: String concatenation ---------- */ void fizzBuzzConcat(int n) { char s[32]; for (int i = 1; i <= n; i++) { s[0] = '\0'; if (i % 3 == 0) strncat(s, "Fizz", sizeof(s) - strlen(s) - 1); if (i % 5 == 0) strncat(s, "Buzz", sizeof(s) - strlen(s) - 1); if (s[0] == '\0') snprintf(s, sizeof(s), "%d", i); printf("%s ", s); } printf("\n"); } /* ---------- Approach 3: Table-driven ---------- */ typedef struct { int divisor; const char *word; } Rule; void fizzBuzzTable(int n, const Rule rules[], int ruleCount) { char s[64]; for (int i = 1; i <= n; i++) { s[0] = '\0'; for (int r = 0; r < ruleCount; r++) { if (i % rules[r].divisor == 0) { strncat(s, rules[r].word, sizeof(s) - strlen(s) - 1); } } if (s[0] == '\0') snprintf(s, sizeof(s), "%d", i); printf("%s ", s); } printf("\n"); } /* ---------- Collector (returns strings) ---------- */ char *fizzBuzzCollector(int n, const Rule rules[], int ruleCount, size_t *outCount) { /* Worst-case: each entry ≤ 63 chars + space */ size_t cap = (size_t)n * 64 + 1; char *result = malloc(cap); if (result == NULL) return NULL; result[0] = '\0'; size_t used = 0; for (int i = 1; i <= n; i++) { char s[64]; s[0] = '\0'; for (int r = 0; r < ruleCount; r++) { if (i % rules[r].divisor == 0) { strncat(s, rules[r].word, sizeof(s) - strlen(s) - 1); } } if (s[0] == '\0') snprintf(s, sizeof(s), "%d", i); size_t len = strlen(s); if (used + len + 2 > cap) break; /* Safety */ memcpy(result + used, s, len); used += len; result[used++] = ' '; result[used] = '\0'; } *outCount = used; return result; } /* ---------- Utility ---------- */ static void printHeader(const char *title) { printf("\n=== %s ===\n", title); } static void demoNaive(void) { printHeader("Naive FizzBuzz (n=20)"); fizzBuzzNaive(20); } static void demoConcat(void) { printHeader("String Concatenation (n=20)"); fizzBuzzConcat(20); } static void demoTable(void) { printHeader("Table-Driven (Fizz/Buzz, n=20)"); Rule rules[] = {{3, "Fizz"}, {5, "Buzz"}}; fizzBuzzTable(20, rules, 2); } static void demoExtended(void) { printHeader("Extended Rules (Fizz/Buzz/Jazz, n=25)"); Rule rules[] = {{3, "Fizz"}, {5, "Buzz"}, {7, "Jazz"}}; fizzBuzzTable(25, rules, 3); } static void demoCollector(void) { printHeader("Collector (returns string)"); Rule rules[] = {{3, "Fizz"}, {5, "Buzz"}}; size_t count = 0; char *result = fizzBuzzCollector(15, rules, 2, &count); if (result == NULL) { printf("Allocation failed\n"); return; } printf("Result: %s\n", result); printf("Length: %zu bytes\n", count); free(result); } static void demoEdgeCases(void) { printHeader("Edge Cases"); printf("n = 0:\n "); fizzBuzzConcat(0); printf("(empty)\n"); printf("\nn = 1:\n "); fizzBuzzConcat(1); printf("\nn = 3:\n "); fizzBuzzConcat(3); printf("\nn = 5:\n "); fizzBuzzConcat(5); printf("\nn = 15 (FizzBuzz):\n "); fizzBuzzConcat(15); printf("\nNegative n = -5:\n "); fizzBuzzConcat(-5); printf("(empty)\n"); } static int readInteger(const char *prompt, int *outValue) { char buffer[MAX_LINE_LEN]; char *endPtr; long val; printf("%s", prompt); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) return -1; errno = 0; val = strtol(buffer, &endPtr, 10); if (endPtr == buffer || (*endPtr != '\n' && *endPtr != '\0')) return -1; if (errno == ERANGE || val < -2147483647 - 1 || val > 2147483647) return -1; *outValue = (int)val; return 0; } static void demoInteractive(void) { int n; if (readInteger("Enter n: ", &n) != 0) { printf("Invalid input.\n"); return; } if (n < 0) { printf("n must be non-negative.\n"); return; } printf("\nTable-driven output:\n "); Rule rules[] = {{3, "Fizz"}, {5, "Buzz"}}; fizzBuzzTable(n, rules, 2); } int main(void) { int choice; char buffer[MAX_LINE_LEN]; printf("=== Fizz Buzz Demo ===\n"); while (1) { printf("\n--- Menu ---\n"); printf("1. Naive (n=20)\n"); printf("2. String Concatenation (n=20)\n"); printf("3. Table-Driven (n=20)\n"); printf("4. Extended Rules Fizz/Buzz/Jazz (n=25)\n"); printf("5. Collector (returns string)\n"); printf("6. Edge Cases\n"); printf("7. Interactive\n"); printf("8. Exit\n"); printf("Enter choice: "); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) { fprintf(stderr, "Error reading choice.\n"); return EXIT_FAILURE; } choice = atoi(buffer); switch (choice) { case 1: demoNaive(); break; case 2: demoConcat(); break; case 3: demoTable(); break; case 4: demoExtended(); break; case 5: demoCollector(); break; case 6: demoEdgeCases(); break; case 7: demoInteractive(); break; case 8: printf("Exiting.\n"); return EXIT_SUCCESS; default: printf("Invalid choice.\n"); break; } } return EXIT_SUCCESS; }

Explanation of Approaches

  1. Naive: Uses explicit branches. Fast and simple; doesn’t scale with more rules.
  2. Concatenation: Appends matching words; scales linearly with rule count.
  3. Table-Driven: Stores rules in an array; adding rules requires no code changes.
  4. Collector: Builds an output string with dynamic memory.
  5. Edge cases: Zero, one, single rules, negatives.

Expected Output (sample run)

text

=== Naive FizzBuzz (n=20) === 1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz === String Concatenation (n=20) === 1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz === Table-Driven (Fizz/Buzz, n=20) === 1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz === Extended Rules (Fizz/Buzz/Jazz, n=25) === 1 2 Fizz 4 Buzz Fizz Jazz 8 Fizz Buzz 11 Fizz 13 Jazz FizzBuzz 16 17 Fizz 19 Buzz FizzJazz 22 23 Fizz Buzz === Collector (returns string) === Result: 1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz Length: 51 bytes === Edge Cases === n = 0: (empty) n = 1: 1 n = 3: 1 2 Fizz n = 5: 1 2 Fizz 4 Buzz n = 15 (FizzBuzz): 1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz Negative n = -5: (empty)

Common Pitfalls, Edge Cases & Memory Considerations

1. Checking Both Conditions Separately

Using i % 3 == 0 && i % 5 == 0 in the naive approach works, but ordering is crucial — the combined case must come first.

c

if (i % 15 == 0) /* FizzBuzz first */ else if (i % 3 == 0) /* Fizz */ else if (i % 5 == 0) /* Buzz */

2. Buffer Overflow in Concatenation

Concatenating multiple words can overflow small buffers.

Solution: Size buffers generously (char s[64]) and use strncat with proper bounds.

c

strncat(s, "Fizz", sizeof(s) - strlen(s) - 1);

3. Using strcat Without Initializing

strcat on uninitialized buffer is undefined behavior.

c

char s[20]; /* BUG: not initialized */ strcat(s, "Fizz"); char s[20] = ""; /* Correct */

4. sprintf Deprecation

sprintf lacks bounds checking.

Solution: Use snprintf.

c

snprintf(s, sizeof(s), "%d", i);

5. Negative n

Loop for (int i = 1; i <= n; i++) naturally skips when n < 0. Document this behavior.

6. Zero n

Empty output is correct.

7. Duplicate Divisors

If two rules share the same divisor, both words are appended (correct). But if divisors are non-coprime, results may surprise.

8. Case Sensitivity

Rule words should be consistent. "Fizz" vs "fizz" matters if compared later.

9. Unicode / Multibyte

strlen, strcat operate on bytes. For UTF-8, still works, but lengths differ from display width.

10. Memory Considerations

  • Naive/concatenation: O(1) extra space.
  • Collector: O(n × word length) — must free.
  • Table: O(k) rule storage.

Summary & Best Practices

Fizz Buzz is a deceptively simple problem that teaches fundamental control flow and scalable design. The table-driven approach is the most extensible and is preferred when rules may grow.

Key Takeaways

  • Naive: simple but doesn’t scale with more rules.
  • Concatenation: appends matching words — linear in rules.
  • Table-driven: clean, extensible, easy to test.
  • Order matters: check combined divisors before individual ones in naive approach.
  • Use snprintf and strncat for safe string operations.
  • Negative n yields empty output.

Best Practices

  1. Prefer the table-driven approach for extensibility.
  2. Use snprintf and strncat for safe string handling.
  3. Initialize string buffers to "".
  4. Size buffers generously (e.g., 64 bytes).
  5. Free dynamically allocated result strings.
  6. Guard against negative or zero n.
  7. Keep rule arrays const when possible.
  8. Separate logic from presentation — collect results, then print.
  9. Test all edge cases: 0, 1, 15, 30.
  10. Document rule ordering semantics.

By mastering Fizz Buzz’s variants, you’ll be better prepared for designing extensible rule-based systems in real projects.

Learn to check if two arrays are equal in C: sorting O(n log n) and hash map O(n) approaches with frequency counting. Complete code examples.
C Programming

Check if Two Arrays Are Equal in C: Complete Guide

Check if Two Arrays Are Equal in C: A Comprehensive Tutorial

Determining whether two arrays are equal—regardless of order but with identical element counts—is a fundamental problem that appears in set comparison, data validation, and caching systems. This tutorial explores two primary approaches: sorting-based comparison (O(n log n)) and hash-based counting (O(n) average).

Conceptual Introduction & Fundamentals

What Does “Arrays Are Equal” Mean?

Two arrays are equal if:

  • Both have the same length.
  • Both contain the same multiset of elements (same elements with same counts).
  • Order does not matter.

Examples:

text

Input: a = [1, 2, 5, 4, 0], b = [2, 4, 5, 0, 1] Output: true Input: a = [1, 2, 5, 4, 0, 2, 1], b = [2, 4, 5, 0, 1, 1, 2] Output: true Input: a = [1, 7, 1], b = [7, 7, 1] Output: false (a has two 1s, b has two 7s)

Why This Problem Matters

  • Multiset comparison: Foundation for set operations with duplicates.
  • Data validation: Verify input arrays match expected.
  • Caching: Check cache-key equivalence.
  • Testing: Assert expected outputs match actual.
  • Interview preparation: Common coding interview question.

Approaches Overview

ApproachTime ComplexitySpace Complexity
SortingO(n log n)O(1) or O(n)
Hashing (frequency)O(n) averageO(n)

Note: Simple linear comparison only works if arrays are already sorted or order must match.

Syntax & Core Mechanics

Approach 1: Sorting

Sort both arrays and compare element by element.

c

#include <stdio.h> #include <stdlib.h> #include <string.h> int compareInts(const void *x, const void *y) { int a = *(const int *)x; int b = *(const int *)y; return (a > b) - (a < b); } int arraysEqualSorted(const int a_in[], int n, const int b_in[], int m) { if (n != m) return 0; int *a = malloc(n * sizeof(int)); int *b = malloc(m * sizeof(int)); if (a == NULL || b == NULL) { free(a); free(b); return 0; } memcpy(a, a_in, n * sizeof(int)); memcpy(b, b_in, m * sizeof(int)); qsort(a, n, sizeof(int), compareInts); qsort(b, m, sizeof(int), compareInts); int result = 1; for (int i = 0; i < n; i++) { if (a[i] != b[i]) { result = 0; break; } } free(a); free(b); return result; } int main(void) { int a[] = {3, 5, 2, 5, 2}; int b[] = {2, 3, 5, 5, 2}; int n = sizeof(a) / sizeof(a[0]); int m = sizeof(b) / sizeof(b[0]); printf("%s\n", arraysEqualSorted(a, n, b, m) ? "true" : "false"); return 0; }

Expected output:

text

true

Time: O(n log n), Space: O(n) with copy, O(1) in-place

Approach 2: Hash Map (Frequency Counting)

Count occurrences in a; decrement counts using b.

c

#include <stdio.h> #include <stdlib.h> #define TABLE_SIZE 8192 typedef struct { int *keys; int *counts; int *occupied; int capacity; } HashMap; static unsigned int hashFunc(int key, int capacity) { return ((unsigned int)key) % (unsigned int)capacity; } HashMap *createHashMap(int capacity) { HashMap *map = malloc(sizeof(HashMap)); if (map == NULL) return NULL; map->keys = malloc(capacity * sizeof(int)); map->counts = calloc(capacity, sizeof(int)); map->occupied = calloc(capacity, sizeof(int)); if (map->keys == NULL || map->counts == NULL || map->occupied == NULL) { free(map->keys); free(map->counts); free(map->occupied); free(map); return NULL; } map->capacity = capacity; return map; } void freeHashMap(HashMap *map) { if (map) { free(map->keys); free(map->counts); free(map->occupied); free(map); } } void increment(HashMap *map, int key) { unsigned int idx = hashFunc(key, map->capacity); int start = (int)idx; while (map->occupied[idx]) { if (map->keys[idx] == key) { map->counts[idx]++; return; } idx = (idx + 1) % (unsigned int)map->capacity; if ((int)idx == start) return; } map->keys[idx] = key; map->counts[idx] = 1; map->occupied[idx] = 1; } int getCount(HashMap *map, int key) { unsigned int idx = hashFunc(key, map->capacity); int start = (int)idx; while (map->occupied[idx]) { if (map->keys[idx] == key) return map->counts[idx]; idx = (idx + 1) % (unsigned int)map->capacity; if ((int)idx == start) break; } return 0; } void decrement(HashMap *map, int key) { unsigned int idx = hashFunc(key, map->capacity); int start = (int)idx; while (map->occupied[idx]) { if (map->keys[idx] == key) { map->counts[idx]--; return; } idx = (idx + 1) % (unsigned int)map->capacity; if ((int)idx == start) return; } } int arraysEqualHash(const int a[], int n, const int b[], int m) { if (n != m) return 0; HashMap *map = createHashMap(TABLE_SIZE); if (map == NULL) return 0; for (int i = 0; i < n; i++) increment(map, a[i]); int result = 1; for (int i = 0; i < m; i++) { int c = getCount(map, b[i]); if (c == 0) { result = 0; break; } decrement(map, b[i]); } freeHashMap(map); return result; } int main(void) { int a[] = {3, 5, 2, 5, 2}; int b[] = {2, 3, 5, 5, 2}; int n = sizeof(a) / sizeof(a[0]); int m = sizeof(b) / sizeof(b[0]); printf("%s\n", arraysEqualHash(a, n, b, m) ? "true" : "false"); return 0; }

Expected output:

text

true

Time: O(n) average, Space: O(n)

Full Practical Code Example with Explanation

This program compares both approaches with verification and edge cases.

c

/** * arrays_equal.c * Checks if two arrays have the same multiset of elements. * Compares sorting and hash-map approaches. */ #include <stdio.h> #include <stdlib.h> #include <string.h> #include <errno.h> #include <time.h> #define MAX_LINE_LEN 100 #define TABLE_SIZE 262144 /* 2^18 */ /* ---------- Approach 1: Sorting ---------- */ static int compareInts(const void *x, const void *y) { int a = *(const int *)x; int b = *(const int *)y; return (a > b) - (a < b); } int arraysEqualSorted(const int a_in[], int n, const int b_in[], int m) { if (n != m) return 0; if (n == 0) return 1; int *a = malloc(n * sizeof(int)); int *b = malloc(m * sizeof(int)); if (a == NULL || b == NULL) { free(a); free(b); return -1; } memcpy(a, a_in, n * sizeof(int)); memcpy(b, b_in, m * sizeof(int)); qsort(a, n, sizeof(int), compareInts); qsort(b, m, sizeof(int), compareInts); int result = 1; for (int i = 0; i < n; i++) { if (a[i] != b[i]) { result = 0; break; } } free(a); free(b); return result; } /* ---------- Approach 2: Hash Map ---------- */ typedef struct { int *keys; int *counts; int *occupied; int capacity; } HashMap; static unsigned int hashFunc(int key, int capacity) { return ((unsigned int)key) % (unsigned int)capacity; } HashMap *createHashMap(int capacity) { HashMap *map = malloc(sizeof(HashMap)); if (map == NULL) return NULL; map->keys = malloc(capacity * sizeof(int)); map->counts = calloc(capacity, sizeof(int)); map->occupied = calloc(capacity, sizeof(int)); if (map->keys == NULL || map->counts == NULL || map->occupied == NULL) { free(map->keys); free(map->counts); free(map->occupied); free(map); return NULL; } map->capacity = capacity; return map; } void freeHashMap(HashMap *map) { if (map) { free(map->keys); free(map->counts); free(map->occupied); free(map); } } void increment(HashMap *map, int key) { unsigned int idx = hashFunc(key, map->capacity); int start = (int)idx; while (map->occupied[idx]) { if (map->keys[idx] == key) { map->counts[idx]++; return; } idx = (idx + 1) % (unsigned int)map->capacity; if ((int)idx == start) return; } map->keys[idx] = key; map->counts[idx] = 1; map->occupied[idx] = 1; } int getCount(HashMap *map, int key) { unsigned int idx = hashFunc(key, map->capacity); int start = (int)idx; while (map->occupied[idx]) { if (map->keys[idx] == key) return map->counts[idx]; idx = (idx + 1) % (unsigned int)map->capacity; if ((int)idx == start) break; } return 0; } void decrement(HashMap *map, int key) { unsigned int idx = hashFunc(key, map->capacity); int start = (int)idx; while (map->occupied[idx]) { if (map->keys[idx] == key) { map->counts[idx]--; return; } idx = (idx + 1) % (unsigned int)map->capacity; if ((int)idx == start) return; } } int arraysEqualHash(const int a[], int n, const int b[], int m) { if (n != m) return 0; if (n == 0) return 1; HashMap *map = createHashMap(TABLE_SIZE); if (map == NULL) return -1; for (int i = 0; i < n; i++) increment(map, a[i]); int result = 1; for (int i = 0; i < m; i++) { int c = getCount(map, b[i]); if (c == 0) { result = 0; break; } decrement(map, b[i]); } freeHashMap(map); return result; } /* ---------- Utility ---------- */ void printArray(const int arr[], int n) { printf("["); for (int i = 0; i < n; i++) { printf("%d", arr[i]); if (i < n - 1) printf(", "); } printf("]"); } static void demoBasic(void) { printf("\n=== Sample Arrays ===\n"); int a1[] = {1, 2, 5, 4, 0}; int b1[] = {2, 4, 5, 0, 1}; int a2[] = {1, 2, 5, 4, 0, 2, 1}; int b2[] = {2, 4, 5, 0, 1, 1, 2}; int a3[] = {1, 7, 1}; int b3[] = {7, 7, 1}; struct { int *a, *b; int n, m; } tests[] = { {a1, b1, 5, 5}, {a2, b2, 7, 7}, {a3, b3, 3, 3} }; for (int t = 0; t < 3; t++) { printf("\nTest %d:\n a = ", t + 1); printArray(tests[t].a, tests[t].n); printf("\n b = "); printArray(tests[t].b, tests[t].m); printf("\n Sorted: %s\n", arraysEqualSorted(tests[t].a, tests[t].n, tests[t].b, tests[t].m) ? "true" : "false"); printf(" Hash: %s\n", arraysEqualHash(tests[t].a, tests[t].n, tests[t].b, tests[t].m) ? "true" : "false"); } printf("\n"); } static void demoEdgeCases(void) { printf("\n=== Edge Cases ===\n"); /* Both empty */ printf("Both empty: sorted=%s, hash=%s\n", arraysEqualSorted(NULL, 0, NULL, 0) ? "true" : "false", arraysEqualHash(NULL, 0, NULL, 0) ? "true" : "false"); /* Different lengths */ int a1[] = {1, 2, 3}; int b1[] = {1, 2}; printf("Different lengths: sorted=%s, hash=%s\n", arraysEqualSorted(a1, 3, b1, 2) ? "true" : "false", arraysEqualHash(a1, 3, b1, 2) ? "true" : "false"); /* Same single element */ int a2[] = {42}; int b2[] = {42}; printf("Single equal: sorted=%s, hash=%s\n", arraysEqualSorted(a2, 1, b2, 1) ? "true" : "false", arraysEqualHash(a2, 1, b2, 1) ? "true" : "false"); /* Different frequencies */ int a3[] = {1, 1, 2}; int b3[] = {1, 2, 2}; printf("Different frequencies: sorted=%s, hash=%s\n", arraysEqualSorted(a3, 3, b3, 3) ? "true" : "false", arraysEqualHash(a3, 3, b3, 3) ? "true" : "false"); /* Negatives */ int a4[] = {-1, -5, -1, 3}; int b4[] = {3, -1, -5, -1}; printf("Negatives: sorted=%s, hash=%s\n", arraysEqualSorted(a4, 4, b4, 4) ? "true" : "false", arraysEqualHash(a4, 4, b4, 4) ? "true" : "false"); /* Zeros */ int a5[] = {0, 0, 1}; int b5[] = {1, 0, 0}; printf("Zeros: sorted=%s, hash=%s\n", arraysEqualSorted(a5, 3, b5, 3) ? "true" : "false", arraysEqualHash(a5, 3, b5, 3) ? "true" : "false"); /* Same array identity */ int a6[] = {5, 4, 3, 2, 1}; printf("Same array: sorted=%s, hash=%s\n", arraysEqualSorted(a6, 5, a6, 5) ? "true" : "false", arraysEqualHash(a6, 5, a6, 5) ? "true" : "false"); } static void performanceTest(void) { printf("\n=== Performance Comparison (n = 100000) ===\n"); int n = 100000; int *a = malloc(n * sizeof(int)); int *b = malloc(n * sizeof(int)); if (a == NULL || b == NULL) { free(a); free(b); return; } srand(42); for (int i = 0; i < n; i++) { a[i] = rand() % 1000; b[i] = a[i]; } /* Shuffle b */ for (int i = n - 1; i > 0; i--) { int j = rand() % (i + 1); int tmp = b[i]; b[i] = b[j]; b[j] = tmp; } clock_t start, end; double t; start = clock(); int r1 = arraysEqualSorted(a, n, b, n); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Sorted: result=%s, time=%.3f ms\n", r1 ? "true" : "false", t); start = clock(); int r2 = arraysEqualHash(a, n, b, n); end = clock(); t = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Hash: result=%s, time=%.3f ms\n", r2 ? "true" : "false", t); printf("Verification: %s\n", (r1 == r2) ? "PASS" : "FAIL"); free(a); free(b); } int main(void) { int choice; char buffer[MAX_LINE_LEN]; printf("=== Arrays Equal Check Demo ===\n"); while (1) { printf("\n--- Menu ---\n"); printf("1. Demonstrate All Approaches\n"); printf("2. Edge Cases\n"); printf("3. Performance Comparison\n"); printf("4. Exit\n"); printf("Enter choice: "); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) { fprintf(stderr, "Error reading choice.\n"); return EXIT_FAILURE; } choice = atoi(buffer); switch (choice) { case 1: demoBasic(); break; case 2: demoEdgeCases(); break; case 3: performanceTest(); break; case 4: printf("Exiting.\n"); return EXIT_SUCCESS; default: printf("Invalid choice.\n"); break; } } return EXIT_SUCCESS; }

Explanation of Approaches

  1. Sorting: Copy inputs (to preserve), sort both, compare element by element.
  2. Hash Map: Count occurrences in a; decrement counts using b. If any b element’s count hits 0 before decrement, arrays differ.

Expected Output (sample run)

text

=== Sample Arrays === Test 1: a = [1, 2, 5, 4, 0] b = [2, 4, 5, 0, 1] Sorted: true Hash: true Test 2: a = [1, 2, 5, 4, 0, 2, 1] b = [2, 4, 5, 0, 1, 1, 2] Sorted: true Hash: true Test 3: a = [1, 7, 1] b = [7, 7, 1] Sorted: false Hash: false === Edge Cases === Both empty: sorted=true, hash=true Different lengths: sorted=false, hash=false Single equal: sorted=true, hash=true Different frequencies: sorted=false, hash=false Negatives: sorted=true, hash=true Zeros: sorted=true, hash=true Same array: sorted=true, hash=true === Performance Comparison (n = 100000) === Sorted: result=true, time=15.234 ms Hash: result=true, time=9.876 ms Verification: PASS

Common Pitfalls, Edge Cases & Memory Considerations

1. Comparing Lengths First

Different lengths → not equal. Always check n == m first.

2. Forgetting Counts

Two arrays with the same set but different counts are not equal.

c

a = [1, 1, 2]; b = [1, 2, 2]; /* Not equal */

3. Mutating Input

Sort-based approach modifies inputs unless copied.

Solution: Copy before sorting.

4. Empty Arrays

Both empty → equal. Guard against NULL arrays.

5. Hash Collisions

Ensure hash table sized appropriately; handle probe loops.

6. Negative Keys

Cast to unsigned int in hash function.

7. Early Exit

Return as soon as mismatch found.

8. Count Overflow

Use int counts; for extreme cases, int64_t.

9. Duplicate Elements

The frequency-counting approach naturally handles duplicates.

10. Memory Considerations

  • Sorted: O(n) copy or O(1) in-place; O(n log n) time.
  • Hash: O(n) extra space; O(n) time.
  • Hash is preferred for large unsorted arrays.

Summary & Best Practices

Checking arrays for multiset equality is a foundational operation. Sorting is simple and memory-efficient; hashing is faster in practice.

Key Takeaways

  • Arrays equal ⟺ same multiset of elements.
  • Sort + compare: O(n log n) time.
  • Hash + count: O(n) average time.
  • Always check length first.
  • Empty arrays are equal.

Best Practices

  1. Check length equality first.
  2. Use hashing for large arrays (O(n) average).
  3. Use sorting when memory is constrained.
  4. Copy inputs before sorting if preservation required.
  5. Use a hash map with frequency counts (not just membership).
  6. Handle negatives via unsigned cast in hash.
  7. Guard against NULL and zero-length arrays.
  8. Free allocated memory.
  9. Test edge cases: empty, single, different counts, negatives.
  10. Profile for large inputs.

By mastering multiset equality checking, you’ll gain practical insight into hash maps, sorting, and array comparison.

Learn to find the unique element where all others appear twice in C: naive, hash map, and optimal XOR approach. Master O(n) time with O(1) space.
C Programming

Unique Number in Array in C: XOR Method Guide

Unique Number in an Array (Every Element Appears Twice Except One) in C

Finding the unique element in an array where every other element appears exactly twice is a classic problem that tests your understanding of bitwise operations and algorithm optimization. This problem demonstrates the elegant power of XOR and is a favorite in coding interviews.

This tutorial provides an in-depth exploration of finding the unique number in C: naive nested-loop approach, hash-map approach, XOR approach, practical implementation, common pitfalls, and best practices.

Conceptual Introduction & Fundamentals

What is the Problem?

Given an array of integers where every element appears exactly twice except for one element that appears only once, find that unique element.

Examples:

text

Input: [2, 3, 5, 4, 5, 3, 4] Output: 2 Input: [2, 2, 5, 5, 20, 30, 30] Output: 20

Why This Problem Matters

  • Bitwise operations: Demonstrates XOR’s cancellation property.
  • Algorithm optimization: O(n²) → O(n) → O(n) with O(1) space.
  • Interview preparation: Extremely common coding interview question.
  • Memory efficiency: XOR approach uses no extra memory.
  • Foundation for advanced problems: Single Number II, III, and related variants.

Key Insight: XOR Cancellation Property

XOR (exclusive OR) has these properties:

PropertyFormula
Identitya ⊕ 0 = a
Self-inversea ⊕ a = 0
Commutativea ⊕ b = b ⊕ a
Associative(a ⊕ b) ⊕ c = a ⊕ (b ⊕ c)

When we XOR all elements, pairs cancel out:

text

2 ⊕ 3 ⊕ 5 ⊕ 4 ⊕ 5 ⊕ 3 ⊕ 4 = 2 ⊕ (3 ⊕ 3) ⊕ (5 ⊕ 5) ⊕ (4 ⊕ 4) = 2 ⊕ 0 ⊕ 0 ⊕ 0 = 2

Approaches Overview

ApproachTime ComplexitySpace Complexity
Naive frequency countO(n²)O(1)
Hash mapO(n)O(n)
XORO(n)O(1)

Syntax & Core Mechanics

Approach 1: Naive Nested Loops

Count occurrences of each element using nested loops.

c

#include <stdio.h> int findUniqueNaive(const int arr[], int n) { for (int i = 0; i < n; i++) { int count = 0; for (int j = 0; j < n; j++) { if (arr[i] == arr[j]) { count++; } } if (count == 1) { return arr[i]; } } return -1; /* Not found */ } int main(void) { int arr[] = {2, 3, 5, 4, 5, 3, 4}; int n = sizeof(arr) / sizeof(arr[0]); printf("Unique element: %d\n", findUniqueNaive(arr, n)); return 0; }

Expected output:

text

Unique element: 2

Time: O(n²), Space: O(1)

Approach 2: Hash Map

Use a hash table to count frequencies, then find the element with count 1.

c

#include <stdio.h> #include <stdlib.h> #define TABLE_SIZE 2048 typedef struct { int *keys; int *counts; int *occupied; int capacity; } HashMap; HashMap *createHashMap(int capacity) { HashMap *map = malloc(sizeof(HashMap)); if (map == NULL) return NULL; map->keys = malloc(capacity * sizeof(int)); map->counts = calloc(capacity, sizeof(int)); map->occupied = calloc(capacity, sizeof(int)); if (map->keys == NULL || map->counts == NULL || map->occupied == NULL) { free(map->keys); free(map->counts); free(map->occupied); free(map); return NULL; } map->capacity = capacity; return map; } void freeHashMap(HashMap *map) { if (map != NULL) { free(map->keys); free(map->counts); free(map->occupied); free(map); } } int hashFunc(int key, int capacity) { return ((key % capacity) + capacity) % capacity; } void increment(HashMap *map, int key) { int idx = hashFunc(key, map->capacity); int start = idx; while (map->occupied[idx]) { if (map->keys[idx] == key) { map->counts[idx]++; return; } idx = (idx + 1) % map->capacity; if (idx == start) return; } map->keys[idx] = key; map->counts[idx] = 1; map->occupied[idx] = 1; } int getCount(HashMap *map, int key) { int idx = hashFunc(key, map->capacity); int start = idx; while (map->occupied[idx]) { if (map->keys[idx] == key) return map->counts[idx]; idx = (idx + 1) % map->capacity; if (idx == start) break; } return 0; } int findUniqueHash(const int arr[], int n) { if (n <= 0) return -1; HashMap *map = createHashMap(TABLE_SIZE); if (map == NULL) return -1; for (int i = 0; i < n; i++) { increment(map, arr[i]); } int result = -1; for (int i = 0; i < n; i++) { if (getCount(map, arr[i]) == 1) { result = arr[i]; break; } } freeHashMap(map); return result; } int main(void) { int arr[] = {2, 3, 5, 4, 5, 3, 4}; int n = sizeof(arr) / sizeof(arr[0]); printf("Unique element: %d\n", findUniqueHash(arr, n)); return 0; }

Expected output:

text

Unique element: 2

Time: O(n) average, Space: O(n)

Approach 3: XOR (Optimal)

XOR all elements; pairs cancel, leaving the unique element.

c

#include <stdio.h> int findUniqueXOR(const int arr[], int n) { int result = 0; for (int i = 0; i < n; i++) { result ^= arr[i]; } return result; } int main(void) { int arr[] = {2, 3, 5, 4, 5, 3, 4}; int n = sizeof(arr) / sizeof(arr[0]); printf("Unique element: %d\n", findUniqueXOR(arr, n)); return 0; }

Expected output:

text

Unique element: 2

Time: O(n), Space: O(1)

Step-by-Step XOR Trace

text

arr = [2, 3, 5, 4, 5, 3, 4] result = 0 result ^= 2 → 2 result ^= 3 → 1 result ^= 5 → 4 result ^= 4 → 0 result ^= 5 → 5 result ^= 3 → 6 result ^= 4 → 2 Final: 2 ✓

Full Practical Code Example with Explanation

This program demonstrates all three approaches with verification, performance comparison, and edge cases.

Program: Unique Element Finder

c

/** * unique_number.c * Finds the unique element where all others appear twice. * Compares naive, hash map, and XOR approaches. */ #include <stdio.h> #include <stdlib.h> #include <string.h> #include <errno.h> #include <time.h> #define MAX_LINE_LEN 100 #define MAX_SIZE 100000 /* Function prototypes */ int findUniqueNaive(const int arr[], int n); int findUniqueHash(const int arr[], int n); int findUniqueXOR(const int arr[], int n); int readInteger(const char *prompt, int *outValue); void printArray(const int arr[], int n); void demonstrateApproaches(void); void performanceComparison(void); void demonstrateEdgeCases(void); int main(void) { int choice; char buffer[MAX_LINE_LEN]; printf("=== Unique Number Finder ===\n"); while (1) { printf("\n--- Menu ---\n"); printf("1. Demonstrate All Approaches\n"); printf("2. Custom Array\n"); printf("3. Performance Comparison\n"); printf("4. Edge Cases\n"); printf("5. Exit\n"); printf("Enter choice: "); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) { fprintf(stderr, "Error reading choice.\n"); return EXIT_FAILURE; } choice = atoi(buffer); switch (choice) { case 1: demonstrateApproaches(); break; case 2: { int arr[MAX_SIZE]; int n; if (readInteger("Enter array size (odd, 1-99999): ", &n) != 0) { fprintf(stderr, "Invalid size.\n"); break; } if (n < 1 || n > MAX_SIZE) { printf("Size must be 1-%d.\n", MAX_SIZE); break; } printf("Enter %d elements:\n", n); for (int i = 0; i < n; i++) { char prompt[50]; snprintf(prompt, sizeof(prompt), " Element %d: ", i); if (readInteger(prompt, &arr[i]) != 0) { fprintf(stderr, "Invalid input.\n"); n = i; break; } } printf("\nArray: "); printArray(arr, n); int xorResult = findUniqueXOR(arr, n); printf("Unique element (XOR): %d\n", xorResult); break; } case 3: performanceComparison(); break; case 4: demonstrateEdgeCases(); break; case 5: printf("Exiting.\n"); return EXIT_SUCCESS; default: printf("Invalid choice. Please try again.\n"); break; } } return EXIT_SUCCESS; } /* Naive: O(n^2) */ int findUniqueNaive(const int arr[], int n) { for (int i = 0; i < n; i++) { int count = 0; for (int j = 0; j < n; j++) { if (arr[i] == arr[j]) count++; } if (count == 1) return arr[i]; } return -1; } /* Hash map: O(n) average */ typedef struct { int *keys; int *counts; int *occupied; int capacity; } HashMap; HashMap *createHashMap(int capacity) { HashMap *map = malloc(sizeof(HashMap)); if (map == NULL) return NULL; map->keys = malloc(capacity * sizeof(int)); map->counts = calloc(capacity, sizeof(int)); map->occupied = calloc(capacity, sizeof(int)); if (map->keys == NULL || map->counts == NULL || map->occupied == NULL) { free(map->keys); free(map->counts); free(map->occupied); free(map); return NULL; } map->capacity = capacity; return map; } void freeHashMap(HashMap *map) { if (map != NULL) { free(map->keys); free(map->counts); free(map->occupied); free(map); } } int hashFunc(int key, int capacity) { return ((key % capacity) + capacity) % capacity; } void increment(HashMap *map, int key) { int idx = hashFunc(key, map->capacity); int start = idx; while (map->occupied[idx]) { if (map->keys[idx] == key) { map->counts[idx]++; return; } idx = (idx + 1) % map->capacity; if (idx == start) return; } map->keys[idx] = key; map->counts[idx] = 1; map->occupied[idx] = 1; } int getCount(HashMap *map, int key) { int idx = hashFunc(key, map->capacity); int start = idx; while (map->occupied[idx]) { if (map->keys[idx] == key) return map->counts[idx]; idx = (idx + 1) % map->capacity; if (idx == start) break; } return 0; } int findUniqueHash(const int arr[], int n) { if (n <= 0) return -1; HashMap *map = createHashMap(n * 2 + 1); if (map == NULL) return -1; for (int i = 0; i < n; i++) { increment(map, arr[i]); } int result = -1; for (int i = 0; i < n; i++) { if (getCount(map, arr[i]) == 1) { result = arr[i]; break; } } freeHashMap(map); return result; } /* XOR: O(n) time, O(1) space */ int findUniqueXOR(const int arr[], int n) { int result = 0; for (int i = 0; i < n; i++) { result ^= arr[i]; } return result; } /* Prints array */ void printArray(const int arr[], int n) { printf("["); for (int i = 0; i < n; i++) { printf("%d", arr[i]); if (i < n - 1) printf(", "); } printf("]\n"); } /* Demonstrates all approaches */ void demonstrateApproaches(void) { printf("\n=== Sample Arrays ===\n"); int arr1[] = {2, 3, 5, 4, 5, 3, 4}; int n1 = sizeof(arr1) / sizeof(arr1[0]); printf("\nArray 1: "); printArray(arr1, n1); printf("Naive: %d\n", findUniqueNaive(arr1, n1)); printf("Hash: %d\n", findUniqueHash(arr1, n1)); printf("XOR: %d\n", findUniqueXOR(arr1, n1)); int arr2[] = {2, 2, 5, 5, 20, 30, 30}; int n2 = sizeof(arr2) / sizeof(arr2[0]); printf("\nArray 2: "); printArray(arr2, n2); printf("Naive: %d\n", findUniqueNaive(arr2, n2)); printf("Hash: %d\n", findUniqueHash(arr2, n2)); printf("XOR: %d\n", findUniqueXOR(arr2, n2)); int arr3[] = {7}; int n3 = sizeof(arr3) / sizeof(arr3[0]); printf("\nArray 3 (single): "); printArray(arr3, n3); printf("XOR: %d\n", findUniqueXOR(arr3, n3)); int arr4[] = {-1, -2, -1, -3, -2}; int n4 = sizeof(arr4) / sizeof(arr4[0]); printf("\nArray 4 (negatives): "); printArray(arr4, n4); printf("XOR: %d\n", findUniqueXOR(arr4, n4)); } /* Performance comparison */ void performanceComparison(void) { printf("\n=== Performance Comparison (n = 100001) ===\n"); int n = 100001; int *arr = malloc(n * sizeof(int)); if (arr == NULL) { fprintf(stderr, "Memory allocation failed.\n"); return; } /* Fill pairs + one unique */ for (int i = 0; i < n / 2; i++) { arr[2 * i] = i + 1; arr[2 * i + 1] = i + 1; } arr[n - 1] = 999999; /* Unique element */ clock_t start, end; double timeTaken; /* Hash */ start = clock(); int hashResult = findUniqueHash(arr, n); end = clock(); timeTaken = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("Hash (O(n)): Result = %d, Time = %.3f ms\n", hashResult, timeTaken); /* XOR */ start = clock(); int xorResult = findUniqueXOR(arr, n); end = clock(); timeTaken = ((double)(end - start)) / CLOCKS_PER_SEC * 1000.0; printf("XOR (O(n)): Result = %d, Time = %.3f ms\n", xorResult, timeTaken); printf("\nVerification: %s\n", (hashResult == xorResult) ? "PASS" : "FAIL"); free(arr); } /* Demonstrates edge cases */ void demonstrateEdgeCases(void) { printf("\n=== Edge Cases ===\n"); /* Case 1: Single element */ int arr1[] = {42}; printf("Case 1: Single element\n"); printf(" Array: "); printArray(arr1, 1); printf(" Unique: %d\n", findUniqueXOR(arr1, 1)); /* Case 2: Unique at start */ int arr2[] = {99, 1, 1, 2, 2}; printf("\nCase 2: Unique at start\n"); printf(" Array: "); printArray(arr2, 5); printf(" Unique: %d\n", findUniqueXOR(arr2, 5)); /* Case 3: Unique at end */ int arr3[] = {1, 1, 2, 2, 99}; printf("\nCase 3: Unique at end\n"); printf(" Array: "); printArray(arr3, 5); printf(" Unique: %d\n", findUniqueXOR(arr3, 5)); /* Case 4: Negative unique */ int arr4[] = {-5, -5, -10, -3, -3}; printf("\nCase 4: Negative unique\n"); printf(" Array: "); printArray(arr4, 5); printf(" Unique: %d\n", findUniqueXOR(arr4, 5)); /* Case 5: Zero unique */ int arr5[] = {1, 1, 0, 2, 2}; printf("\nCase 5: Zero unique\n"); printf(" Array: "); printArray(arr5, 5); printf(" Unique: %d\n", findUniqueXOR(arr5, 5)); /* Case 6: Large unique value */ int arr6[] = {1, 1, 2147483647, 2, 2}; printf("\nCase 6: Large unique value\n"); printf(" Array: "); printArray(arr6, 5); printf(" Unique: %d\n", findUniqueXOR(arr6, 5)); /* Case 7: All pairs, no unique (invalid input) */ int arr7[] = {1, 1, 2, 2}; printf("\nCase 7: Invalid input (all pairs)\n"); printf(" Array: "); printArray(arr7, 4); printf(" XOR result: %d (no unique exists)\n", findUniqueXOR(arr7, 4)); } /* Reads an integer safely */ int readInteger(const char *prompt, int *outValue) { char buffer[MAX_LINE_LEN]; char *endPtr; long val; printf("%s", prompt); fflush(stdout); if (fgets(buffer, sizeof(buffer), stdin) == NULL) { return -1; } errno = 0; val = strtol(buffer, &endPtr, 10); if (endPtr == buffer || (*endPtr != '\n' && *endPtr != '\0')) { return -1; } if (errno == ERANGE || val < -2147483647 - 1 || val > 2147483647) { return -1; } *outValue = (int)val; return 0; }

Explanation of Approaches

  1. Naive: Counts occurrences with nested loops; O(n²).
  2. Hash: Uses a hash table to count frequencies; O(n) average.
  3. XOR: XORs all elements; pairs cancel leaving the unique; O(n), O(1) space.

Expected Output (sample run)

text

=== Unique Number Finder === --- Menu --- 1. Demonstrate All Approaches 2. Custom Array 3. Performance Comparison 4. Edge Cases 5. Exit Enter choice: 1 === Sample Arrays === Array 1: [2, 3, 5, 4, 5, 3, 4] Naive: 2 Hash: 2 XOR: 2 Array 2: [2, 2, 5, 5, 20, 30, 30] Naive: 20 Hash: 20 XOR: 20 Array 3 (single): [7] XOR: 7 Array 4 (negatives): [-1, -2, -1, -3, -2] XOR: -3 --- Menu --- ... Enter choice: 3 === Performance Comparison (n = 100001) === Hash (O(n)): Result = 999999, Time = 18.456 ms XOR (O(n)): Result = 999999, Time = 0.312 ms Verification: PASS --- Menu --- ... Enter choice: 4 === Edge Cases === Case 1: Single element Array: [42] Unique: 42 Case 2: Unique at start Array: [99, 1, 1, 2, 2] Unique: 99 Case 3: Unique at end Array: [1, 1, 2, 2, 99] Unique: 99 Case 4: Negative unique Array: [-5, -5, -10, -3, -3] Unique: -10 Case 5: Zero unique Array: [1, 1, 0, 2, 2] Unique: 0 Case 6: Large unique value Array: [1, 1, 2147483647, 2, 2] Unique: 2147483647 Case 7: Invalid input (all pairs) Array: [1, 1, 2, 2] XOR result: 0 (no unique exists)

Common Pitfalls, Edge Cases & Memory Considerations

1. Assumption of Exactly One Unique

XOR returns 0 if all elements are paired (no unique). The problem guarantees exactly one unique.

2. Empty Array

XOR returns 0 for empty arrays, which may be misinterpreted as a valid unique value.

c

if (n <= 0) return -1; /* Handle explicitly */

3. Integer Overflow

XOR does not overflow; it operates bit-by-bit.

4. Negative Values

XOR works correctly with negative numbers (two’s complement).

c

int arr[] = {-1, -2, -1, -3, -2}; /* XOR = -3 */

5. Hash Table Sizing

Hash table should be sized ≥ 2× array size for good performance.

c

createHashMap(n * 2 + 1);

6. Hash Table Overflow

If the table fills up, insertions silently fail.

Solution: Size the table generously (2× or more).

7. Using unordered_map in C++

C does not have built-in hash maps; you must implement or use uthash.

Solution: Use XOR when possible (simplest and fastest).

8. Memory Leaks

Free hash table memory after use.

c

freeHashMap(map);

9. Large Unique Value

Unique value can be any int; use int not unsigned.

10. Memory Considerations

  • Naive: O(1) extra space.
  • Hash: O(n) extra space.
  • XOR: O(1) extra space — preferred.
  • XOR is the most efficient and simplest approach.

Summary & Best Practices

Finding the unique element where all others appear twice is elegantly solved using XOR. The XOR of all elements cancels pairs, leaving only the unique element.

Key Takeaways

  • XOR cancels identical pairs: a ⊕ a = 0.
  • XOR identity: a ⊕ 0 = a.
  • XOR all elements → unique element.
  • Naive: O(n²) time, O(1) space.
  • Hash: O(n) time, O(n) space.
  • XOR: O(n) time, O(1) space — optimal.
  • Problem guarantees exactly one unique element.

Best Practices

  1. Use XOR for O(n) time and O(1) space.
  2. Verify the problem’s guarantee (exactly one unique).
  3. Handle empty arrays explicitly.
  4. Test with negative values and zeros.
  5. Use hash map only when XOR isn’t applicable (e.g., unique appears 3 times).
  6. Free hash map memory after use.
  7. Size hash tables generously for good performance.
  8. Use int (signed) for the result.
  9. Document the XOR property in comments.
  10. Profile for large arrays.

By mastering this problem, you’ll understand the power of XOR and be prepared for related bitwise problems like Single Number II and III.