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

Solution 1: Simulation

We can directly simulate to calculate the XOR result of all elements in the array.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

Python3

class Solution:
    def xorOperation(self, n: int, start: int) -> int:
        return reduce(xor, ((start + 2 * i) for i in range(n)))

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;
    }
}

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

TypeScript

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