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.

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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.


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Constraints:


	1 &lt;= n &lt;= 1000
	0 &lt;= start &lt;= 1000
	n == nums.length


class Solution:
    def xorOperation(self, n: int, start: int) -> int:
        nums = [start + 2*i for i in range(n)] #generate list of numbers
        ans = nums[0]
        for i in range(1,n):
            ans = ans^nums[i] # XOR operation
        return ans
        
        
class Solution {
    public int xorOperation(int n, int start) {
        int nums[]=new int[n];
        for(int i=0;i<n;i++)
            nums[i] = start + 2 * i;
        for(int i=1;i<n;i++)
            nums[i] = nums[i-1]^nums[i];
        return nums[n-1];
    }
}
class Solution {
public:
    int xorOperation(int n, int start) {

        int res = start;

        for(int i=1 ; i<n ; i++){
            res = res ^ (start + 2 * i);
        }

        return res;
    }
};
var xorOperation = function(n, start) {
    let arr = []
    for(let i=0; i<n; i++){
        arr.push(start+2*i)
    }
    return arr.reduce((a,c)=> a^c)
};
XOR Operation in an Array