Skip to content

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 \(O(1)\).

1
2
3
class Solution:
    def xorOperation(self, n: int, start: int) -> int:
        return reduce(xor, ((start + 2 * i) for i in range(n)))

1
2
3
4
5
6
7
8
9
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
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;
    }
};

1
2
3
4
5
6
func xorOperation(n int, start int) (ans int) {
    for i := 0; i < n; i++ {
        ans ^= start + 2*i
    }
    return
}

1
2
3
4
5
6
7
function xorOperation(n: number, start: number): number {
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        ans ^= start + 2 * i;
    }
    return ans;
}

Comments