// 1879. Minimum XOR Sum of Two Arrays #include "leetcode.h" /* * you are given two integer arrays 'nums1' and 'nums2' of length 'n'. the xor * sum of the two integer arrays is '(nums1[0] ^ nums2[0]) + (nums1[1] ^ * nums2[1]) + ...' (0-indexed). rearrange the elements of 'nums2' such that * resulting xor sum is minimised. return the xor sum after the rearrangement. */ static int dp[1 << 14] = {[0 ... 16383] = INT_MAX}; int dfs(int *nums1, int *nums2, int n, int i, int mask) { if (i >= n) return 0; if (dp[mask] == -1) for (int j = 0; j < n; ++j) if (!(mask & (1 << j))) dp[mask] = fmin(dp[mask], (nums1[i] ^ nums2[j]) + dfs(nums1, nums2, n, i + 1, mask + (1 << j))); return dp[mask]; } int minimumXORSum(int *nums1, int nums1Size, int *nums2, int nums2Size) { return dfs(nums1, nums2, nums1Size, 0, 0); } int main() { int n11[] = {1, 2}, n21[] = {2, 3}, n12[] = {1, 0, 3}, n22[] = {5, 3, 4}; printf("%d\n", minimumXORSum(n11, ARRAY_SIZE(n11), n21, ARRAY_SIZE(n21))); // expect: 2 printf("%d\n", minimumXORSum(n12, ARRAY_SIZE(n12), n22, ARRAY_SIZE(n22))); // expect: 8 }