// 1787. Make the XOR of All Segments Equal to Zero #include "leetcode.h" /* * given an array 'nums' and an integer 'k'. the xor of a segment '[left, * right]' where 'left <= right' is the xor of all the elements with indices * between 'left' and 'right', inclusive 'nums[left] xor nums[left + 1] xor ... * xor nums[right]'. return the minimum number of elements to change in the * array such that the 'xor' of all segments of size 'k' is equal to zero */ int minChanges(int *nums, int numsSize, int k) { int freq[k][1025], dp[k + 1][1025]; memset(freq, 0, sizeof(freq)); for (int i = 0; i <= k; i++) for (int j = 0; j < 1025; j++) dp[i][j] = 1e9 + 5; dp[0][0] = 0; for (int i = 0; i < numsSize; i++) freq[i % k][nums[i]]++; int prev_min = 0, prev_min2 = 1e9 + 5; for (int i = 0; i < k; i++) { for (int j = 0; j < 1025; j++) for (int l = 0; l < numsSize; l += k) if ((j ^ nums[l]) < 1025) dp[i + 1][j ^ nums[l]] = fmin( dp[i + 1][j ^ nums[l]], dp[i][j] + ceil((numsSize - i) * 1.0 / k) - freq[i][nums[l]]); for (int j = 0; j < 1025; j++) { dp[i + 1][j] = fmin(dp[i + 1][j], ceil((numsSize - i) * 1.0 / k) + prev_min); prev_min2 = fmin(prev_min2, dp[i + 1][j]); } prev_min = prev_min2; } return dp[k][0]; } int main() { int n1[] = {1, 2, 0, 3, 0}, n2[] = {3, 4, 5, 2, 1, 7, 3, 4, 7}, n3[] = {1, 2, 4, 1, 2, 5, 1, 2, 6}; printf("%d\n", minChanges(n1, ARRAY_SIZE(n1), 1)); // expect: 3 printf("%d\n", minChanges(n2, ARRAY_SIZE(n2), 3)); // expect: 3 printf("%d\n", minChanges(n3, ARRAY_SIZE(n3), 3)); // expect: 3 }