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Formatted question description: https://leetcode.ca/all/2425.html

# 2425. Bitwise XOR of All Pairings

• Difficulty: Medium.
• Related Topics: .
• Similar Questions: .

## Problem

You are given two 0-indexed arrays, nums1 and nums2, consisting of non-negative integers. There exists 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 <= 10^5

• 0 <= nums1[i], nums2[j] <= 10^9

## Solution (Java, C++, Python)

• class Solution {
public int xorAllNums(int[] nums1, int[] nums2) {
int ans = 0;
if (nums2.length % 2 == 1) {
for (int v : nums1) {
ans ^= v;
}
}
if (nums1.length % 2 == 1) {
for (int v : nums2) {
ans ^= v;
}
}
return ans;
}
}

• class Solution {
public:
int xorAllNums(vector<int>& nums1, vector<int>& nums2) {
int ans = 0;
if (nums2.size() % 2 == 1) {
for (int v : nums1) {
ans ^= v;
}
}
if (nums1.size() % 2 == 1) {
for (int v : nums2) {
ans ^= v;
}
}
return ans;
}
};

• class Solution:
def xorAllNums(self, nums1: List[int], nums2: List[int]) -> int:
ans = 0
if len(nums2) & 1:
for v in nums1:
ans ^= v
if len(nums1) & 1:
for v in nums2:
ans ^= v
return ans


• func xorAllNums(nums1 []int, nums2 []int) int {
ans := 0
if len(nums2)%2 == 1 {
for _, v := range nums1 {
ans ^= v
}
}
if len(nums1)%2 == 1 {
for _, v := range nums2 {
ans ^= v
}
}
return ans
}

• function xorAllNums(nums1: number[], nums2: number[]): number {
let ans = 0;
if (nums2.length % 2 != 0) {
ans ^= nums1.reduce((a, c) => a ^ c, 0);
}
if (nums1.length % 2 != 0) {
ans ^= nums2.reduce((a, c) => a ^ c, 0);
}
return ans;
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).