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// 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
}
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