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, 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(n² * 2ⁿ), where n is the length of the input arrays. Storage complexity: O(2ⁿ).

Code

    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)