diff options
Diffstat (limited to 'min_xor_sum_two_arr.c')
-rw-r--r-- | min_xor_sum_two_arr.c | 35 |
1 files changed, 35 insertions, 0 deletions
diff --git a/min_xor_sum_two_arr.c b/min_xor_sum_two_arr.c new file mode 100644 index 0000000..492a0f8 --- /dev/null +++ b/min_xor_sum_two_arr.c @@ -0,0 +1,35 @@ +// 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 +} |