Problem Description

The problem asks to find the minimum XOR sum of two arrays nums1 and nums2 after rearranging the elements of nums2. The XOR sum is defined as the sum of element-wise XOR operations between the two arrays. The length of both arrays is n, where 1 <= n <= 14.

Solution Explanation

The constraints (1 <= n <= 14) indicate that a brute-force approach checking all permutations of nums2 is computationally feasible. However, a more efficient solution uses dynamic programming and bit manipulation.

The core idea is to represent the subsets of nums2 using bitmasks. A bitmask of length n can represent all possible subsets: a 1 at position i indicates that nums2[i] is included in the subset. The dynamic programming approach then iterates through the elements of nums1 and builds up the minimum XOR sum for all possible subsets of nums2.

State: dp[i][mask] represents the minimum XOR sum achievable using the first i elements of nums1 and a subset of nums2 represented by the bitmask mask.

Transition: For each element nums1[i], we iterate through all possible subsets of nums2 (mask). If a bit is set in mask (meaning the corresponding element from nums2 is included), we can calculate the XOR sum with nums1[i] and update the dp[i+1][mask] value accordingly.

Base Case: dp[0][0] = 0 (no elements from nums1 or nums2 used).

Final Result: dp[n][(1 << n) - 1] (using all elements from both arrays).

Code Explanation (Python3 - Solution 1)

class Solution:
    def minimumXORSum(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums2)
        f = [[inf] * (1 << n) for _ in range(n + 1)]  # Initialize DP table
        f[0][0] = 0  # Base case
 
        for i, x in enumerate(nums1, 1):  # Iterate through nums1
            for j in range(1 << n):  # Iterate through all subsets of nums2
                for k in range(n):  # Iterate through elements of nums2
                    if j >> k & 1:  # Check if k-th bit is set in j (nums2[k] is in subset)
                        f[i][j] = min(f[i][j], f[i - 1][j ^ (1 << k)] + (x ^ nums2[k]))
                        # Update DP table: min(current value, previous value + XOR sum)
        return f[-1][-1]  # Final result

Time Complexity: O(n * 2n * n) = O(n2 * 2n). The outer loops iterate through nums1 and all subsets of nums2, while the inner loop checks each element in nums2.

Space Complexity: O(n * 2n) to store the DP table.

Code Explanation (Python3 - Solution 2)

This solution optimizes space by using a 1D DP array. The outer loop iterates backward to avoid overwriting DP values.

class Solution:
    def minimumXORSum(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums2)
        f = [inf] * (1 << n)
        f[0] = 0
        for x in nums1:
            for j in range((1 << n) - 1, -1, -1):
                for k in range(n):
                    if j >> k & 1:
                        f[j] = min(f[j], f[j ^ (1 << k)] + (x ^ nums2[k]))
        return f[-1]

Time Complexity: Same as Solution 1: O(n2 * 2n)

Space Complexity: O(2n), significantly reduced from the 2D approach.

Code Explanation (Python3 - Solution 3)

This version further refines the approach by iterating through subsets in a more structured way, reducing redundant calculations.

class Solution:
    def minimumXORSum(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums2)
        f = [inf] * (1 << n)
        f[0] = 0
        for i in range(1, 1 << n):
            k = i.bit_count() - 1
            for j in range(n):
                if i >> j & 1:
                    f[i] = min(f[i], f[i ^ (1 << j)] + (nums1[k] ^ nums2[j]))
        return f[-1]

Time Complexity: Still O(n2 * 2n), but with potential performance improvements due to the iteration order.

Space Complexity: O(2n)

The provided solutions in Java, C++, Go, and TypeScript follow the same logic and complexity analysis as the Python solutions. They differ only in syntax and specific library functions used for bit manipulation and DP table initialization.