1486. XOR Operation in an Array

Problem

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

Solution (Java)

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;
        }
        int result = 0;
        for (int num : nums) {
            result ^= num;
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).