# Solving Leetcode Interviews in Seconds with AI: Bitwise XOR of All Pairings


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2425" 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 0-indexed arrays, nums1 and nums2, consisting of non-negative integers. Let there be another array, nums3, which contains the bitwise XOR of all pairings of integers between nums1 and nums2 (every integer in nums1 is paired with every integer in nums2 exactly once). Return the bitwise XOR of all integers in nums3.   Example 1:  Input: nums1 = [2,1,3], nums2 = [10,2,5,0] Output: 13 Explanation: A possible nums3 array is [8,0,7,2,11,3,4,1,9,1,6,3]. The bitwise XOR of all these numbers is 13, so we return 13.  Example 2:  Input: nums1 = [1,2], nums2 = [3,4] Output: 0 Explanation: All possible pairs of bitwise XORs are nums1[0] ^ nums2[0], nums1[0] ^ nums2[1], nums1[1] ^ nums2[0], and nums1[1] ^ nums2[1]. Thus, one possible nums3 array is [2,5,1,6]. 2 ^ 5 ^ 1 ^ 6 = 0, so we return 0.    Constraints:  1 <= nums1.length, nums2.length <= 105 0 <= nums1[i], nums2[j] <= 109  

	# Explanation
	Here's the approach:

*   The XOR operation is associative and commutative. This means we can rearrange the order of XOR operations without affecting the result.
*   `(a ^ b) ^ (a ^ c) = b ^ c`. More generally, `(a ^ x1) ^ (a ^ x2) ^ ... ^ (a ^ xn) = a ^ (a ^ ... ^ a) ^ (x1 ^ x2 ^ ... ^ xn)` where `a` appears `n` times.  If `n` is even, the `a` terms cancel out completely due to XOR properties (a^a = 0). If n is odd, they reduce to just `a`.
*   Therefore, we can XOR all elements of `nums1` together, XOR all elements of `nums2` together, and then XOR the results if the product of array lengths is odd.

*   **Time Complexity:** O(n + m), where n is the length of `nums1` and m is the length of `nums2`.  **Space Complexity:** O(1).

	
	# Code
	```python
	def get_xor_sum(nums1, nums2):
    """
    Calculates the XOR sum of all pairings between two arrays.

    Args:
        nums1: The first array of non-negative integers.
        nums2: The second array of non-negative integers.

    Returns:
        The bitwise XOR of all integers in the array of XOR pairings.
    """

    xor_nums1 = 0
    for num in nums1:
        xor_nums1 ^= num

    xor_nums2 = 0
    for num in nums2:
        xor_nums2 ^= num

    return xor_nums1 & xor_nums2
	```