Introduction

In this blog post, we will explore how to solve the LeetCode problem "1563" 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

There are several stones arranged in a row, and each stone has an associated value which is an integer given in the array stoneValue. In each round of the game, Alice divides the row into two non-empty rows (i.e. left row and right row), then Bob calculates the value of each row which is the sum of the values of all the stones in this row. Bob throws away the row which has the maximum value, and Alice's score increases by the value of the remaining row. If the value of the two rows are equal, Bob lets Alice decide which row will be thrown away. The next round starts with the remaining row. The game ends when there is only one stone remaining. Alice's is initially zero. Return the maximum score that Alice can obtain. Example 1: Input: stoneValue = [6,2,3,4,5,5] Output: 18 Explanation: In the first round, Alice divides the row to [6,2,3], [4,5,5]. The left row has the value 11 and the right row has value 14. Bob throws away the right row and Alice's score is now 11. In the second round Alice divides the row to [6], [2,3]. This time Bob throws away the left row and Alice's score becomes 16 (11 + 5). The last round Alice has only one choice to divide the row which is [2], [3]. Bob throws away the right row and Alice's score is now 18 (16 + 2). The game ends because only one stone is remaining in the row. Example 2: Input: stoneValue = [7,7,7,7,7,7,7] Output: 28 Example 3: Input: stoneValue = [4] Output: 0 Constraints: 1 <= stoneValue.length <= 500 1 <= stoneValue[i] <= 106

Explanation

Here's the solution to the problem, with explanations and code:

• Approach: Dynamic programming is used to solve this problem. dp[i][j] stores the maximum score Alice can obtain from the subarray stoneValue[i...j]. The transitions consider all possible splits of the subarray [i...j], calculate the sums of the left and right subarrays, and update dp[i][j] based on the minimum sum (Alice's score) and the maximum score achievable from the remaining subarray.

• Complexity:

  • Runtime: O(n^3), where n is the length of stoneValue.
  • Storage: O(n^2)

Code

    def stoneGameV(stoneValue):
    n = len(stoneValue)
    if n <= 1:
        return 0

    prefix_sum = [0] * (n + 1)
    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + stoneValue[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
            for k in range(i, j):
                left_sum = prefix_sum[k + 1] - prefix_sum[i]
                right_sum = prefix_sum[j + 1] - prefix_sum[k + 1]

                if left_sum < right_sum:
                    dp[i][j] = max(dp[i][j], left_sum + dp[i][k])
                elif left_sum > right_sum:
                    dp[i][j] = max(dp[i][j], right_sum + dp[k + 1][j])
                else:
                    dp[i][j] = max(dp[i][j], left_sum + max(dp[i][k], dp[k + 1][j]))

    return dp[0][n - 1]