# Solving Leetcode Interviews in Seconds with AI: Divisor Game


	# 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](https://www.chatmagic.app), 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
	```python
	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)
	```