Introduction

In this blog post, we will explore how to solve the LeetCode problem "1140" 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 continue their games with piles of stones. There are a number of piles arranged in a row, and each pile has a positive integer number of stones piles[i]. The objective of the game is to end with the most stones. Alice and Bob take turns, with Alice starting first. On each player's turn, that player can take all the stones in the first X remaining piles, where 1 <= X <= 2M. Then, we set M = max(M, X). Initially, M = 1. The game continues until all the stones have been taken. Assuming Alice and Bob play optimally, return the maximum number of stones Alice can get. Example 1: Input: piles = [2,7,9,4,4] Output: 10 Explanation: If Alice takes one pile at the beginning, Bob takes two piles, then Alice takes 2 piles again. Alice can get 2 + 4 + 4 = 10 stones in total. If Alice takes two piles at the beginning, then Bob can take all three piles left. In this case, Alice get 2 + 7 = 9 stones in total. So we return 10 since it's larger. Example 2: Input: piles = [1,2,3,4,5,100] Output: 104 Constraints: 1 <= piles.length <= 100 1 <= piles[i] <= 104

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: We use dynamic programming to store the maximum score Alice can obtain starting from a given index with a given maximum move M.
• Top-Down (Memoization): We implement the DP approach using recursion with memoization to avoid redundant calculations.

• State Definition: The state dp[i][m] represents the maximum stones Alice can collect when the remaining piles start from index i and the maximum allowed move is m.

• Time Complexity: O(n² * m), where n is the number of piles and m is the maximum possible value of M (bounded by n).

• Space Complexity: O(n²), dominated by the memoization table.

Code

    def stoneGameII(piles):
    n = len(piles)
    prefix_sum = [0] * (n + 1)
    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + piles[i]

    dp = {}  # Memoization table: dp[(i, m)] = max Alice's score

    def solve(i, m):
        if i >= n:
            return 0
        if (i, m) in dp:
            return dp[(i, m)]

        max_score = 0
        for x in range(1, 2 * m + 1):
            if i + x > n:
                break
            score = prefix_sum[n] - prefix_sum[i] - solve(i + x, max(m, x)) # Alice's score + Bob's optimal score from the remaining subproblem
            max_score = max(max_score, score)

        dp[(i, m)] = max_score
        return max_score

    return solve(0, 1)