1690. Stone Game VII

Description

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 i^th 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

Solutions

Solution 1: Memorization Search

First, we preprocess to get the prefix sum array $s$, where $s[i]$ represents the total sum of the first $i$ stones.

Next, we design a function $dfs(i, j)$, which represents the score difference between the first and second players when the remaining stones are $stones[i], stones[i + 1], \dots, stones[j]$. The answer is $dfs(0, n - 1)$.

The calculation process of the function $dfs(i, j)$ is as follows:

If $i > j$, it means there are no stones currently, so return $0$;
Otherwise, the first player has two choices, which are to remove $stones[i]$ or $stones[j]$, and then calculate the score difference, i.e., $a = s[j + 1] - s[i + 1] - dfs(i + 1, j)$ and $b = s[j] - s[i] - dfs(i, j - 1)$. We take the larger of the two as the return value of $dfs(i, j)$.

During the process, we use memorization search, i.e., use an array $f$ to record the return value of the function $dfs(i, j)$, to avoid repeated calculations.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the number of stones.

Python3JavaC++GoTypeScript

class Solution:
    def stoneGameVII(self, stones: List[int]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i > j:
                return 0
            a = s[j + 1] - s[i + 1] - dfs(i + 1, j)
            b = s[j] - s[i] - dfs(i, j - 1)
            return max(a, b)

        s = list(accumulate(stones, initial=0))
        ans = dfs(0, len(stones) - 1)
        dfs.cache_clear()
        return ans

class Solution {
    private int[] s;
    private Integer[][] f;

    public int stoneGameVII(int[] stones) {
        int n = stones.length;
        s = new int[n + 1];
        f = new Integer[n][n];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + stones[i];
        }
        return dfs(0, n - 1);
    }

    private int dfs(int i, int j) {
        if (i > j) {
            return 0;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        int a = s[j + 1] - s[i + 1] - dfs(i + 1, j);
        int b = s[j] - s[i] - dfs(i, j - 1);
        return f[i][j] = Math.max(a, b);
    }
}

class Solution {
public:
    int stoneGameVII(vector<int>& stones) {
        int n = stones.size();
        int f[n][n];
        memset(f, 0, sizeof f);
        int s[n + 1];
        s[0] = 0;
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + stones[i];
        }
        function<int(int, int)> dfs = [&](int i, int j) {
            if (i > j) {
                return 0;
            }
            if (f[i][j]) {
                return f[i][j];
            }
            int a = s[j + 1] - s[i + 1] - dfs(i + 1, j);
            int b = s[j] - s[i] - dfs(i, j - 1);
            return f[i][j] = max(a, b);
        };
        return dfs(0, n - 1);
    }
};

func stoneGameVII(stones []int) int {
    n := len(stones)
    s := make([]int, n+1)
    f := make([][]int, n)
    for i, x := range stones {
        s[i+1] = s[i] + x
        f[i] = make([]int, n)
    }
    var dfs func(int, int) int
    dfs = func(i, j int) int {
        if i > j {
            return 0
        }
        if f[i][j] != 0 {
            return f[i][j]
        }
        a := s[j+1] - s[i+1] - dfs(i+1, j)
        b := s[j] - s[i] - dfs(i, j-1)
        f[i][j] = max(a, b)
        return f[i][j]
    }
    return dfs(0, n-1)
}

function stoneGameVII(stones: number[]): number {
    const n = stones.length;
    const s: number[] = Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] + stones[i];
    }
    const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
    const dfs = (i: number, j: number): number => {
        if (i > j) {
            return 0;
        }
        if (f[i][j]) {
            return f[i][j];
        }
        const a = s[j + 1] - s[i + 1] - dfs(i + 1, j);
        const b = s[j] - s[i] - dfs(i, j - 1);
        return (f[i][j] = Math.max(a, b));
    };
    return dfs(0, n - 1);
}

Solution 2: Dynamic Programming

We can convert the memoization search in Solution 1 into dynamic programming. We define $f[i][j]$ as the score difference between the first and second players when the remaining stones are $stones[i], stones[i + 1], \dots, stones[j]$. Therefore, the answer is $f[0][n - 1]$.

The state transition equation is as follows:

$$ f[i][j] = \max(s[j + 1] - s[i + 1] - f[i + 1][j], s[j] - s[i] - f[i][j - 1]) $$

When calculating $f[i][j]$, we need to ensure that $f[i + 1][j]$ and $f[i][j - 1]$ have been calculated. Therefore, we need to enumerate $i$ in descending order and $j$ in ascending order.

Finally, the answer is $f[0][n - 1]$.