1486. XOR Operation in an Array

EasyLeetCode

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

How to solve the problem

For loop to traverse to n-1, XOR the element with the current result.

Code

class Solution:
    def xorOperation(self, n: int, start: int) -> int:
        num = 0
        for i in range(n):
            num ^= start + 2 * i
        return num

Complexity

Time complexity: O(n)

Traversal is needed.

Space complexity: O(1)

No extra data structures are used.