Formatted question description: https://leetcode.ca/all/1879.html

1879. Minimum XOR Sum of Two Arrays

Level

Hard

Description

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] <= 10^7

Solution

Use dynamic programming with compressed states. Create a 2D array dp of n rows and 2^n columns, where dp[i][j] represents the minimum XOR sum of the first i + 1 numbers in nums1 when the combination in nums2 is j. Finally, return dp[n - 1][2^n - 1].

class Solution {
    public int minimumXORSum(int[] nums1, int[] nums2) {
        int length = nums1.length;
        int[][] dp = new int[length][1 << length];
        for (int i = 0; i < length; i++)
            dp[0][1 << i] = nums1[0] ^ nums2[i];
        for (int i = 1; i < length; i++) {
            Arrays.fill(dp[i], Integer.MAX_VALUE);
            for (int j = 1; j < 1 << length; j++) {
                int remain = j;
                while (remain > 0) {
                    int curBit = remain & (-remain);
                    if ((j & curBit) != 0) {
                        int k = (int) (Math.log(curBit) / Math.log(2));
                        dp[i][j] = Math.min(dp[i][j], dp[i - 1][j - curBit] + (nums1[i] ^ nums2[k]));
                    }
                    remain -= curBit;
                }
            }
        }
        return dp[length - 1][(1 << length) - 1];
    }
}

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