LeetCode #810 — HARD

Chalkboard XOR Game

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an array of integers nums represents the numbers written on a chalkboard.

Alice and Bob take turns erasing exactly one number from the chalkboard, with Alice starting first. If erasing a number causes the bitwise XOR of all the elements of the chalkboard to become 0, then that player loses. The bitwise XOR of one element is that element itself, and the bitwise XOR of no elements is 0.

Also, if any player starts their turn with the bitwise XOR of all the elements of the chalkboard equal to 0, then that player wins.

Return true if and only if Alice wins the game, assuming both players play optimally.

Example 1:

Input: nums = [1,1,2]
Output: false
Explanation: 
Alice has two choices: erase 1 or erase 2. 
If she erases 1, the nums array becomes [1, 2]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 2 = 3. Now Bob can remove any element he wants, because Alice will be the one to erase the last element and she will lose. 
If Alice erases 2 first, now nums become [1, 1]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 1 = 0. Alice will lose.

Example 2:

Input: nums = [0,1]
Output: true

Example 3:

Input: nums = [1,2,3]
Output: true

Constraints:

1 <= nums.length <= 1000
0 <= nums[i] < 2¹⁶

Step 01

Brute Force Baseline

Problem summary: You are given an array of integers nums represents the numbers written on a chalkboard. Alice and Bob take turns erasing exactly one number from the chalkboard, with Alice starting first. If erasing a number causes the bitwise XOR of all the elements of the chalkboard to become 0, then that player loses. The bitwise XOR of one element is that element itself, and the bitwise XOR of no elements is 0. Also, if any player starts their turn with the bitwise XOR of all the elements of the chalkboard equal to 0, then that player wins. Return true if and only if Alice wins the game, assuming both players play optimally.

Baseline thinking

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

Pattern signal: Array · Math · Bit Manipulation

Example 1

[1,1,2]

Example 2

[0,1]

Example 3

[1,2,3]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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

Largest constraint values

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 #810: Chalkboard XOR Game
class Solution {
    public boolean xorGame(int[] nums) {
        return nums.length % 2 == 0 || Arrays.stream(nums).reduce(0, (a, b) -> a ^ b) == 0;
    }
}

// Accepted solution for LeetCode #810: Chalkboard XOR Game
func xorGame(nums []int) bool {
	if len(nums)%2 == 0 {
		return true
	}
	x := 0
	for _, v := range nums {
		x ^= v
	}
	return x == 0
}

# Accepted solution for LeetCode #810: Chalkboard XOR Game
class Solution:
    def xorGame(self, nums: List[int]) -> bool:
        return len(nums) % 2 == 0 or reduce(xor, nums) == 0

// Accepted solution for LeetCode #810: Chalkboard XOR Game
struct Solution;

impl Solution {
    fn xor_game(nums: Vec<i32>) -> bool {
        let xor = nums.iter().fold(0, |xor, x| xor ^ x);
        xor == 0 || nums.len() % 2 == 0
    }
}

#[test]
fn test() {
    let nums = vec![1, 1, 2];
    let res = false;
    assert_eq!(Solution::xor_game(nums), res);
}

// Accepted solution for LeetCode #810: Chalkboard XOR Game
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #810: Chalkboard XOR Game
// class Solution {
//     public boolean xorGame(int[] nums) {
//         return nums.length % 2 == 0 || Arrays.stream(nums).reduce(0, (a, b) -> a ^ b) == 0;
//     }
// }

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.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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.