# 1486. XOR Operation in an Array

## Description

You are given an integer n and an integer start.

Define an array nums where nums[i] = start + 2 * i (0-indexed) and n == nums.length.

Return the bitwise XOR of all elements of nums.

Example 1:

Input: n = 5, start = 0
Output: 8
Explanation: Array nums is equal to [0, 2, 4, 6, 8] where (0 ^ 2 ^ 4 ^ 6 ^ 8) = 8.
Where "^" corresponds to bitwise XOR operator.


Example 2:

Input: n = 4, start = 3
Output: 8
Explanation: Array nums is equal to [3, 5, 7, 9] where (3 ^ 5 ^ 7 ^ 9) = 8.


Constraints:

• 1 <= n <= 1000
• 0 <= start <= 1000
• n == nums.length

## Solutions

• class Solution {
public int xorOperation(int n, int start) {
int ans = 0;
for (int i = 0; i < n; ++i) {
ans ^= start + 2 * i;
}
return ans;
}
}

• class Solution {
public:
int xorOperation(int n, int start) {
int ans = 0;
for (int i = 0; i < n; ++i) {
ans ^= start + 2 * i;
}
return ans;
}
};

• class Solution:
def xorOperation(self, n: int, start: int) -> int:
ans = 0
for i in range(n):
ans ^= start + 2 * i
return ans


• func xorOperation(n int, start int) (ans int) {
for i := 0; i < n; i++ {
ans ^= start + 2*i
}
return
}

• function xorOperation(n: number, start: number): number {
let ans = 0;
for (let i = 0; i < n; ++i) {
ans ^= start + 2 * i;
}
return ans;
}