Introduction

In this blog post, we will explore how to solve the LeetCode problem "1690" 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. There are n stones arranged in a row. On each player's turn, they can remove either the leftmost stone or the rightmost stone from the row and receive points equal to the sum of the remaining stones' values in the row. The winner is the one with the higher score when there are no stones left to remove. Bob found that he will always lose this game (poor Bob, he always loses), so he decided to minimize the score's difference. Alice's goal is to maximize the difference in the score. Given an array of integers stones where stones[i] represents the value of the ith stone from the left, return the difference in Alice and Bob's score if they both play optimally. Example 1: Input: stones = [5,3,1,4,2] Output: 6 Explanation: - Alice removes 2 and gets 5 + 3 + 1 + 4 = 13 points. Alice = 13, Bob = 0, stones = [5,3,1,4]. - Bob removes 5 and gets 3 + 1 + 4 = 8 points. Alice = 13, Bob = 8, stones = [3,1,4]. - Alice removes 3 and gets 1 + 4 = 5 points. Alice = 18, Bob = 8, stones = [1,4]. - Bob removes 1 and gets 4 points. Alice = 18, Bob = 12, stones = [4]. - Alice removes 4 and gets 0 points. Alice = 18, Bob = 12, stones = []. The score difference is 18 - 12 = 6. Example 2: Input: stones = [7,90,5,1,100,10,10,2] Output: 122 Constraints: n == stones.length 2 <= n <= 1000 1 <= stones[i] <= 1000

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Dynamic Programming: Use dynamic programming to store the maximum score difference Alice can achieve for each subproblem (subarray of stones).

• State Definition: dp[i][j] represents the maximum score difference Alice can achieve when playing optimally on the subarray stones[i:j+1].

• Recursion: Alice maximizes the difference, while Bob minimizes it. The base case is when only one stone is left.

• Time Complexity: O(n^2), where n is the number of stones.

• Space Complexity: O(n^2)

Code

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

    dp = [[0] * n for _ in range(n)]

    for length in range(2, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            dp[i][j] = max(prefix_sum[j + 1] - prefix_sum[i + 1] - dp[i + 1][j],
                           prefix_sum[j] - prefix_sum[i] - dp[i][j - 1])

    return dp[0][n - 1]