Introduction

In this blog post, we will explore how to solve the LeetCode problem "1510" 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 are n stones in a pile. On each player's turn, that player makes a move consisting of removing any non-zero square number of stones in the pile. Also, if a player cannot make a move, he/she loses the game. Given a positive integer n, return true if and only if Alice wins the game otherwise return false, assuming both players play optimally. Example 1: Input: n = 1 Output: true Explanation: Alice can remove 1 stone winning the game because Bob doesn't have any moves. Example 2: Input: n = 2 Output: false Explanation: Alice can only remove 1 stone, after that Bob removes the last one winning the game (2 -> 1 -> 0). Example 3: Input: n = 4 Output: true Explanation: n is already a perfect square, Alice can win with one move, removing 4 stones (4 -> 0). Constraints: 1 <= n <= 105

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: We use dynamic programming to store the results of subproblems. dp[i] stores whether the first player wins starting with i stones.

• Base Case: dp[0] is false because if there are no stones, the first player loses.

• Recursive Relation: A player wins if there exists a square number s such that removing s stones leads to the opponent losing.

• Time Complexity & Storage Complexity: O(n * sqrt(n)) time complexity, O(n) space complexity.

Code

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

    Args:
        n: The initial number of stones in the pile.

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

    dp = [False] * (n + 1)

    for i in range(1, n + 1):
        j = 1
        while j * j <= i:
            if not dp[i - j * j]:
                dp[i] = True
                break
            j += 1

    return dp[n]