LeetCode in Kotlin

1486. XOR Operation in an Array

Easy

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

class Solution {
    fun xorOperation(n: Int, start: Int): Int {
        val nums = IntArray(n)
        for (i in 0 until n) {
            nums[i] = start + 2 * i
        }
        var result = 0
        for (num in nums) {
            result = result xor num
        }
        return result
    }
}

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