How do you solve Chalkboard XOR Game on LeetCode?

Chalkboard XOR Game (LeetCode #810) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: If the total XOR is 0, Alice wins immediately.

What is the brute force solution for Chalkboard XOR Game?

The brute force approach for Chalkboard XOR Game is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we simulate each possible move Alice can make and check if it leads to a win or loss. This involves calculating the XOR for every possible state of the chalkboard after each player's turn. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Chalkboard XOR Game?

Chalkboard XOR Game uses the following concepts: Array, Math, Bit Manipulation, Brainteaser, Game Theory. The recommended patterns to study are: Game Theory, Bit Manipulation.

What are the interview tips for Chalkboard XOR Game?

Always think about the properties of the operations involved (like XOR). Consider edge cases, such as very small arrays. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions.

What are common mistakes when solving Chalkboard XOR Game?

Not considering the properties of XOR and how they affect the game. Failing to recognize that the length of the array can change the outcome.

#810

Chalkboard XOR Game

Hard

ArrayMathBit ManipulationBrainteaserGame TheoryGame TheoryBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the properties of XOR. If the XOR of all numbers is 0, Alice wins immediately. If the XOR is non-zero, the outcome depends on the number of elements: if it's even, Alice will win; if odd, Bob will win.

⚙️

Algorithm

3 steps

1Step 1: Calculate the XOR of all elements in nums.
2Step 2: If the XOR is 0, return true (Alice wins).
3Step 3: If the XOR is non-zero, check the length of nums: if it's even, return true (Alice wins); if odd, return false (Bob wins).

solution.py5 lines

1def xorGame(nums):
2    total_xor = 0
3    for num in nums:
4        total_xor ^= num
5    return total_xor == 0 or len(nums) % 2 == 0

ℹ

Complexity note: The time complexity is O(n) because we only need to iterate through the array once to compute the XOR. The space complexity is O(1) as we are using a constant amount of space.

1If the total XOR is 0, Alice wins immediately.
2If the total XOR is non-zero, the length of the array determines the winner.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.