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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;
    }
    
    

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