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 i = 1, result = 0;
        while(i<=n){
            result ^= start;
            start+=2;
            i++;
        }
        return result;
    }
}

Here, we are going to solve using using XOR operator. In the problem, they clearly mentioned, that start increased by 2 and also we need to iterate till n numbers from start.

1) Initialize i as 1, result as 0.
2) Iterate i till n times, while iterating, each time increase start by 2, i by 1 and do XOR operation with result and again store it in result.
3) Finally return result.


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