LeetCode-in-Java

1879. Minimum XOR Sum of Two Arrays

Hard

You are given two integer arrays nums1 and nums2 of length n.

The XOR sum of the two integer arrays is (nums1[0] XOR nums2[0]) + (nums1[1] XOR nums2[1]) + ... + (nums1[n - 1] XOR nums2[n - 1]) (0-indexed).

• For example, the XOR sum of [1,2,3] and [3,2,1] is equal to (1 XOR 3) + (2 XOR 2) + (3 XOR 1) = 2 + 0 + 2 = 4.

Rearrange the elements of nums2 such that the resulting XOR sum is minimized.

Return the XOR sum after the rearrangement.

Example 1:

Input: nums1 = [1,2], nums2 = [2,3]

Output: 2

Explanation: Rearrange nums2 so that it becomes [3,2]. The XOR sum is (1 XOR 3) + (2 XOR 2) = 2 + 0 = 2.

Example 2:

Input: nums1 = [1,0,3], nums2 = [5,3,4]

Output: 8

Explanation: Rearrange nums2 so that it becomes [5,4,3]. The XOR sum is (1 XOR 5) + (0 XOR 4) + (3 XOR 3) = 4 + 4 + 0 = 8.

Constraints:

• n == nums1.length
• n == nums2.length
• 1 <= n <= 14
• 0 <= nums1[i], nums2[i] <= 107

Solution

import java.util.Arrays;

public class Solution {
    public int minimumXORSum(int[] nums1, int[] nums2) {
        int l = nums1.length;
        int[] dp = new int[1 << l];
        Arrays.fill(dp, -1);
        dp[0] = 0;
        return dfs(dp.length - 1, l, nums1, nums2, dp, l);
    }

    private int dfs(int state, int length, int[] nums1, int[] nums2, int[] dp, int totalLength) {
        if (dp[state] >= 0) {
            return dp[state];
        }
        int min = Integer.MAX_VALUE;
        int currIndex = totalLength - length;
        int i = 0;
        for (int index = 0; i < length; index++) {
            if (((state >> index) & 1) == 1) {
                int result = dfs(state ^ (1 << index), length - 1, nums1, nums2, dp, totalLength);
                min = Math.min(min, (nums2[currIndex] ^ nums1[index]) + result);
                i++;
            }
        }
        dp[state] = min;
        return min;
    }
}

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