Formatted question description: https://leetcode.ca/all/1835.html

# 1835. Find XOR Sum of All Pairs Bitwise AND

Hard

## Description

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  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 = , arr2 = 

Output: 4

Explanation: The list = [12 AND 4] = . The XOR sum = 4.

Constraints:

• 1 <= arr1.length, arr2.length <= 10^5
• 0 <= arr1[i], arr2[j] <= 10^9

## Solution

Use the distributive law. For integers a, b and c, there is (a ^ b) & c == (a & c) ^ (b & c) and a & (b ^ c) = (a & b) ^ (a & c).

For 0 <= i < arr1.length, the XOR sum of arr1[i] & arr2[j] where 0 <= j < arr2.length is equal to arr1[i] & (arr2 ^ arr2 ^ ... ^ arr2[arr2.length - 1]).

Therefore, the XOR sum of the list is calculated as follows.

(arr1 ^ arr1 ^ ... ^ arr1[arr1.length - 1]) & (arr2 ^ arr2 ^ ... ^ arr2[arr2.length - 1])

class Solution {
public int getXORSum(int[] arr1, int[] arr2) {
int xor1 = 0, xor2 = 0;
for (int num : arr1)
xor1 ^= num;
for (int num : arr2)
xor2 ^= num;
return xor1 & xor2;
}
}