LeetCode 1486: XOR Operation in an Array

Meta Data

#

Difficulty: easy First Attempt: 2025-12-18

• Submissions: Accepted (Python) → Accepted (C++) → Wrong Answer (C++) → Accepted (C++)

Approach

#

The i-th number is start + 2 * i, and n is at most 1000, so XOR them one by one.

My Attempts

#

Both Python and C++ passed first; rewriting the C++ later, I typed n * i instead of 2 * i. For n = 5, start = 0 the answer is 0 ^ 2 ^ 4 ^ 6 ^ 8 = 8; that version computed 0 ^ 5 ^ 10 ^ 15 ^ 20 = 20.

Solution

#

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

Complexity

#

Time O(n), space O(1).

Mistakes to Avoid

#

• The step is a fixed 2, not n.

Interview Explanation

#

Generate each start + 2i by definition and accumulate the XOR in one variable. n is small so O(n) is enough; for O(1) there is the periodic formula for 1 ^ 2 ^ ... ^ k.