text: /*
  XOR Operation in an Array
  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.
  Example 3:
  
  Input: n = 1, start = 7
  Output: 7
  Example 4:
  
  Input: n = 10, start = 5
  Output: 2
   
  
  Constraints:
  
  1 <= n <= 1000
  0 <= start <= 1000
  n == nums.length
  */
  
  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;
      }
  };