Stone Game Vii - Leetcode Solution

Problem Statement:

Alice and Bob take turns playing a game, with Alice starting first.

There is a pile of n stones, and each player has a score. On each player's turn, they can remove a stone from the pile and add its value to their score. The player who has the highest score at the end, wins.

Bob found that he will always lose this game (i.e., his score will always be less than Alice's score) no matter how he plays. So, Bob decides to minimize the score's difference. Alice's goal is to maximize the difference in the score.

Given an array of n 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.

Solution:

To solve this problem, we need to first understand the given information correctly. We will maintain two scores, Alice's score and Bob's score, and our goal is to find the difference between them.

As per the problem statement, Alice goes first, and both players play optimally to maximize or minimize the difference as appropriate. So, we need to check what will happen if Alice takes the ith stone and Bob takes the jth stone.

If Alice takes the ith stone and Bob takes the jth stone, then the difference in score would be calculated as:

score_diff = score(Alice) - score(Bob) = stones[i] + score_diff(i+1, j) - score(Bob) + stones[j] + score_diff(i, j-1) - score(Alice)

Here, we have called recurively for the ith+1 to jth position and i-th to j-1th position, as these are the two remaining piles once Alice and Bob have picked up the stones from ith and jth positions respectively. Since both players are playing optimally, we would want to maximize the difference in each step. Thus, we will return the maximum value of score_diff for each i, j combination.

But, if we implement the above approach without any optimization, we will be recalcualting the score difference for many same i,j pairs over and over again, thus leading to time complexity of O(n^3), which is not good for larger inputs. We need to optimize the above approach.

We can observe that for the i and j position, we just need to consider the i+1 and j-1 positions for calculating score_diff(i,j) as it will handle all other cases.

Thus, we can maintain a prefix sum of the array and use this to calculate the score difference for all i's and j's in O(1) time. The prefix sum can be calculated in O(n) time and stored for later use.

Once we have the prefix sum and the two limits i and j, we can directly use them to calculate the score_diff for the remaining piles using the prefix sum. Hence, the time complexity will be reduced from O(n^3) to O(n^2).

We can also optimize our approach further by using dynamic programming. So, we can use a dp[i][j] array to store the optimal maximum difference for the i and j pile. And we can use the previously calculated dp table values to calculate the score_diff for the remaining piles, leading to the time complexity of O(n^2).

Let's now take a look at the final solution.

Final Solution:

A detailed solution of the Stone Game VII problem on leetcode using Python is given below:

def stoneGameVII(stones: List[int]) -> int: n = len(stones)

# calculate prefix sum
prefixSum = [0] * (n + 1)
for i in range(n):
    prefixSum[i+1] = prefixSum[i] + stones[i]

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

# fill the dp table
for i in range(n-2, -1, -1):
    for j in range(i+1, n):
        scoreDiff1 = prefixSum[j+1] - prefixSum[i+1] - dp[i+1][j]
        scoreDiff2 = prefixSum[j] - prefixSum[i] - dp[i][j-1]
        dp[i][j] = max(scoreDiff1, scoreDiff2)

return dp[0][n-1]

This solution has a time complexity of O(n^2) and a space complexity of O(n^2).

We first calculate the prefix sum of the stone array, iterate over all i and j limits of piles. Then, we get the scoreDiff for each pile by substracting the dp table of adjacent piles. Then we get the maximum of the scoreDiff of two piles. Store the maximum value in the dp table and return the top-right corner of the dp table.

Conclusion:

In this problem, we learned about using dynamic programming to solve complex game theory problems. The problem challenges the engineers to find the possible optimal strategy to change the course of the game and maximize the profit. We looked at the detailed solution of the Stone Game VII problem on leetcode using Python. We utilized prefix-sum, memoization and dynamic programming at the end to optimize the time complexity and minimize repeated calculations for the ith,jth piles. These game theory problems are complex and need the developers to analyze the situation thoroughly to find the optimal strategy.

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