# Solving Leetcode Interviews in Seconds with AI: Minimum XOR Sum of Two Arrays


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1879" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the solution to minimize the XOR sum after rearranging `nums2`:

*   **Key Idea:** The problem can be solved using Dynamic Programming with bitmasking. The core idea is to try every possible permutation of `nums2` and keep track of the minimum XOR sum encountered so far.
*   **State Representation:** `dp[mask]` represents the minimum XOR sum achievable by using the elements of `nums2` corresponding to the bits set in `mask`.
*   **Transition:** For each element in `nums1`, we iterate through the unused elements in `nums2` and compute the XOR sum, updating the `dp` table accordingly.

*   **Runtime Complexity:** O(n<sup>2</sup> * 2<sup>n</sup>), where n is the length of the input arrays. Storage complexity: O(2<sup>n</sup>).

	
	# Code
	```python
	def minimum_xor_sum(nums1, nums2):
    """
    Calculates the minimum XOR sum of two arrays after rearranging nums2.

    Args:
        nums1: The first list of integers.
        nums2: The second list of integers.

    Returns:
        The minimum XOR sum achievable after rearranging nums2.
    """
    n = len(nums1)
    dp = {}

    def solve(index, mask):
        if index == n:
            return 0

        if (index, mask) in dp:
            return dp[(index, mask)]

        min_xor = float('inf')
        for i in range(n):
            if (mask >> i) & 1 == 0:
                new_mask = mask | (1 << i)
                current_xor = (nums1[index] ^ nums2[i]) + solve(index + 1, new_mask)
                min_xor = min(min_xor, current_xor)

        dp[(index, mask)] = min_xor
        return min_xor

    return solve(0, 0)
	```