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XOR Operation in an Array

EasyUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

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.

Examples

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

Solution

Method 1 – Simple Iterative XOR

Intuition

The problem asks for the XOR of a sequence where each element is start + 2*i. XOR is associative and commutative, so we can compute the result by iterating through the sequence and applying XOR step by step.

Approach

  1. Initialize ans = 0.
  2. For i from 0 to n-1:
    • Compute val = start + 2*i.
    • ans ^= val.
  3. Return ans.

Code

C++
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;
    }
};
Go
func xorOperation(n int, start int) int {
    ans := 0
    for i := 0; i < n; i++ {
        ans ^= start + 2*i
    }
    return ans
}
Java
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;
    }
}
Kotlin
class Solution {
    fun xorOperation(n: Int, start: Int): Int {
        var ans = 0
        for (i in 0 until n) {
            ans = ans xor (start + 2 * i)
        }
        return ans
    }
}
Python
class Solution:
    def xorOperation(self, n: int, start: int) -> int:
        ans = 0
        for i in range(n):
            ans ^= start + 2 * i
        return ans
Rust
impl Solution {
    pub fn xor_operation(n: i32, start: i32) -> i32 {
        let mut ans = 0;
        for i in 0..n {
            ans ^= start + 2 * i;
        }
        ans
    }
}
TypeScript
class Solution {
    xorOperation(n: number, start: number): number {
        let ans = 0;
        for (let i = 0; i < n; i++) {
            ans ^= start + 2 * i;
        }
        return ans;
    }
}

Complexity

  • ⏰ Time complexity: O(n) — We compute n values and XOR them.
  • 🧺 Space complexity: O(1) — Only a constant amount of extra space is used.

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