1879. Minimum XOR Sum of Two Arrays

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] <= 107

 

Solutions

Solution: Dynamic Programming + Bit Manipulation

• Time complexity: O(n*2ⁿ)
• Space complexity: O(2ⁿ)

 

JavaScript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
const minimumXORSum = function (nums1, nums2) {
  const n = nums1.length;
  const dp = Array.from({ length: 1 << n }, () => Number.MAX_SAFE_INTEGER);

  const sumXor = (index, mask) => {
    if (index >= n) return 0;
    if (dp[mask] !== Number.MAX_SAFE_INTEGER) return dp[mask];
    const num = nums1[index];
    let result = Number.MAX_SAFE_INTEGER;

    for (let bit = 0; bit < n; bit++) {
      if (mask & (1 << bit)) continue;
      const xor = num ^ nums2[bit];
      const sum = xor + sumXor(index + 1, mask | (1 << bit));

      result = Math.min(sum, result);
    }

    dp[mask] = result;

    return result;
  };

  return sumXor(0, 0);
};