How do you solve Divisor Game on LeetCode?

Divisor Game (LeetCode #1025) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Alice wins if n is even; Bob wins if n is odd.

What is the brute force solution for Divisor Game?

The brute force approach for Divisor Game is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we simulate every possible move Alice and Bob can make. We recursively explore all valid moves until we reach a base case where a player cannot make a move, determining the winner based on that. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Divisor Game?

Divisor Game uses the following concepts: Math, Dynamic Programming, Brainteaser, Game Theory. The recommended patterns to study are: Game Theory, Dynamic Programming.

What are the interview tips for Divisor Game?

Always start by analyzing the problem for patterns or properties (like parity). Think about edge cases and how they affect the game. Be prepared to explain your reasoning clearly, especially if you simplify the solution.

What are common mistakes when solving Divisor Game?

Not recognizing the importance of even vs. odd in determining the winner. Overcomplicating the recursive approach instead of simplifying to a parity check.

#1025

Divisor Game

Easy

MathDynamic ProgrammingBrainteaserGame TheoryGame TheoryDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(1)Space O(1)

The optimal solution leverages the observation that if n is even, Alice can always make it odd for Bob, leading to a win. Conversely, if n is odd, Bob can always return it to even for Alice, leading to his win. Thus, the outcome depends solely on the parity of n.

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Algorithm

3 steps

1Step 1: Check if n is even.
2Step 2: If n is even, return true (Alice wins).
3Step 3: If n is odd, return false (Bob wins).

solution.py2 lines

1def divisorGame(n):
2    return n % 2 == 0

ℹ

Complexity note: The time complexity is O(1) because we only perform a single modulus operation, and space complexity is also O(1) as we use no additional space.

1Alice wins if n is even; Bob wins if n is odd.
2The game can be reduced to a simple parity check.

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