# Solving Leetcode Interviews in Seconds with AI: Find XOR Sum of All Pairs Bitwise AND


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1835" 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
	> The XOR sum of a list is the bitwise XOR of all its elements. If the list only contains one element, then its XOR sum will be equal to this element.  For example, the XOR sum of [1,2,3,4] is equal to 1 XOR 2 XOR 3 XOR 4 = 4, and the XOR sum of [3] is equal to 3.  You are given two 0-indexed arrays arr1 and arr2 that consist only of non-negative integers. Consider the list containing the result of arr1[i] AND arr2[j] (bitwise AND) for every (i, j) pair where 0 <= i < arr1.length and 0 <= j < arr2.length. Return the XOR sum of the aforementioned list.   Example 1:  Input: arr1 = [1,2,3], arr2 = [6,5] Output: 0 Explanation: The list = [1 AND 6, 1 AND 5, 2 AND 6, 2 AND 5, 3 AND 6, 3 AND 5] = [0,1,2,0,2,1]. The XOR sum = 0 XOR 1 XOR 2 XOR 0 XOR 2 XOR 1 = 0.  Example 2:  Input: arr1 = [12], arr2 = [4] Output: 4 Explanation: The list = [12 AND 4] = [4]. The XOR sum = 4.    Constraints:  1 <= arr1.length, arr2.length <= 105 0 <= arr1[i], arr2[j] <= 109  

	# Explanation
	Here's the solution to the problem:

*   **Key Insight:**  The XOR operation has the property that `(a & b) ^ (a & c) == a & (b ^ c)`.  We can use this distributive property to simplify the overall XOR sum calculation. Specifically, the XOR sum of all `arr1[i] & arr2[j]` can be rewritten as XORing `arr1[i] & (XOR of all elements in arr2)`.
*   **Reduce to Single XOR:** Compute the XOR sum of `arr2` first. Then, iterate through `arr1`, computing the AND of each element in `arr1` with the XOR sum of `arr2`, and XORing the results.
*   **Efficiency:** This approach avoids generating the entire list of `arr1[i] & arr2[j]` values, which would be O(n*m) in size.

*   **Complexity:** Time Complexity: O(n + m), where n is the length of arr1 and m is the length of arr2.  Space Complexity: O(1).

	
	# Code
	```python
	def get_xor_sum(arr1: list[int], arr2: list[int]) -> int:
    """
    Calculates the XOR sum of the list containing the result of arr1[i] AND arr2[j]
    for every (i, j) pair.

    Args:
        arr1: The first list of non-negative integers.
        arr2: The second list of non-negative integers.

    Returns:
        The XOR sum of the list.
    """

    xor_sum_arr2 = 0
    for num in arr2:
        xor_sum_arr2 ^= num

    result = 0
    for num in arr1:
        result ^= (num & xor_sum_arr2)

    return result
	```