LeetCode #1486 — EASY

XOR Operation in an Array

Build confidence with an intuition-first walkthrough focused on math fundamentals.

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The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Bit Manipulation

Example 1

5
0

Example 2

4
3

Step 02

Core Insight

What unlocks the optimal approach

Simulate the process, create an array nums and return the Bitwise XOR of all elements of it.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1486: XOR Operation in an Array
class Solution {
    public int xorOperation(int n, int start) {
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans ^= start + 2 * i;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1486: XOR Operation in an Array
func xorOperation(n int, start int) (ans int) {
	for i := 0; i < n; i++ {
		ans ^= start + 2*i
	}
	return
}

# Accepted solution for LeetCode #1486: XOR Operation in an Array
class Solution:
    def xorOperation(self, n: int, start: int) -> int:
        return reduce(xor, ((start + 2 * i) for i in range(n)))

// Accepted solution for LeetCode #1486: XOR Operation in an Array
struct Solution;

impl Solution {
    fn xor_operation(n: i32, start: i32) -> i32 {
        (0..n).fold(0, |acc, i| acc ^ (start + 2 * i))
    }
}

#[test]
fn test() {
    let n = 5;
    let start = 0;
    let res = 8;
    assert_eq!(Solution::xor_operation(n, start), res);
    let n = 4;
    let start = 3;
    let res = 8;
    assert_eq!(Solution::xor_operation(n, start), res);
    let n = 1;
    let start = 7;
    let res = 7;
    assert_eq!(Solution::xor_operation(n, start), res);
    let n = 10;
    let start = 5;
    let res = 2;
    assert_eq!(Solution::xor_operation(n, start), res);
}

// Accepted solution for LeetCode #1486: XOR Operation in an Array
function xorOperation(n: number, start: number): number {
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        ans ^= start + 2 * i;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.