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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

• Java
• C++
• Python
• Go
• TypeScript

• 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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