// 1442. Count Triplets That Can Form Two Arrays of Equal XOR #include "leetcode.h" /* * given an array of integers 'arr'. we want to select three indices 'i', 'j', * and 'k' where '(0 <= i < j <= k < arr.length)'. lets define 'a' and 'b' as * follows. 'a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]', 'b = arr[j] ^ arr[j + * 1] ^ ... ^ arr[k]'. note that '^' denotes the bitwise xor operation. return * the number of triplets (i, j, and k) where 'a == b' */ int countTriplets(int *arr, int arrSize) { if (arrSize < 2) return 0; int ans = 0; for (int i = 0; i < arrSize; i++) { int tmp = arr[i]; for (int j = i + 1; j < arrSize; j++) { tmp ^= arr[j]; if (!tmp) ans += j - i; } } return ans; } int main() { int a1[] = {2, 3, 1, 6, 7}, a2[] = {1, 1, 1, 1, 1}; printf("%d\n", countTriplets(a1, ARRAY_SIZE(a1))); // expect: 4 printf("%d\n", countTriplets(a2, ARRAY_SIZE(a2))); // expect: 10 }