Introduction

In this blog post, we will explore how to solve the LeetCode problem "1025" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Alice and Bob take turns playing a game, with Alice starting first. Initially, there is a number n on the chalkboard. On each player's turn, that player makes a move consisting of: Choosing any x with 0 < x < n and n % x == 0. Replacing the number n on the chalkboard with n - x. Also, if a player cannot make a move, they lose the game. Return true if and only if Alice wins the game, assuming both players play optimally. Example 1: Input: n = 2 Output: true Explanation: Alice chooses 1, and Bob has no more moves. Example 2: Input: n = 3 Output: false Explanation: Alice chooses 1, Bob chooses 1, and Alice has no more moves. Constraints: 1 <= n <= 1000

Explanation

Here's the approach to solve this game problem, along with the Python code:

• Dynamic Programming (Memoization): Use dynamic programming to store the results of subproblems. dp[i] will store whether the first player wins starting with the number i.
• Base Case: If there are no valid moves (no x exists), the current player loses, so dp[i] is False.

• Recursive Relation: A player wins if there exists a move x such that the opponent loses after that move.

• Runtime Complexity: O(n^2), Storage Complexity: O(n)

Code

    def divisorGame(n: int) -> bool:
    """
    Determines if Alice wins the divisor game, assuming both players play optimally.

    Args:
        n: The initial number on the chalkboard.

    Returns:
        True if Alice wins, False otherwise.
    """

    dp = [None] * (n + 1)  # dp[i] stores whether the first player wins starting with number i

    def solve(num):
        if num <= 1:
            return False

        if dp[num] is not None:
            return dp[num]

        for x in range(1, num):
            if num % x == 0:
                if not solve(num - x):  # If the opponent loses after this move
                    dp[num] = True
                    return True

        dp[num] = False  # If no winning move is found
        return False

    return solve(n)