LeetCode : 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

class Solution {
    public int xorOperation(int n, int start) {
        
        int[] ar = new int[n];
        for(int i = 0; i < n; i++) {
            ar[i] = start + 2*i;
        }

        int result = start;
        for(int i = 0; i < n-1; i++) {
            result = result ^ ar[i+1];
        }
        return result;  
    }
}

int result = start;
for(int i = 0; i < n-1; i++) {
result = result ^ ar[i+1];
}
return result;