Formatted question description: https://leetcode.ca/all/1690.html

# 1690. Stone Game VII (Medium)

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

Related Topics:
Dynamic Programming

Similar Questions:

## Solution 1. Bottom-up DP

Let dp[i][j] be the maximum difference the first player can get if the players play on A[i..j].

dp[i][j] = max(
sum(i + 1, j) - dp[i + 1][j],   // if the first player choose A[i]
sum(i, j - 1) - dp[i][j - 1]    // if the first player choose A[j]
)


where sum(i, j) is A[i] + ... + A[j]. We can get sum(i, j) using prefix sum array.

Explanation of the DP formula

Each player needs to play optimally to get as many points as possible and make the other player get as less as possible. So the game is actually the same for both of them.

After Alice finshes the her first round, ignoring the points Alice made there, the game to Bob is exactly the same as if he is the first player.

So each round on A[i..j], no matter who plays first, the first player always has two options:

1. pick A[i], the first player get sum(i + 1, j) points, and we need to deduct the maximum point difference the next player can get in the remaining game, i.e. dp[i + 1][j]
2. pick A[j], the first player get sum(i, j - 1) points, and we need to deduct the maximum point difference the next player can get in the remaining game, i.e. dp[i][j - 1]

And the first player simply pick the option more advantageous to him/her.

// OJ: https://leetcode.com/contest/weekly-contest-219/problems/stone-game-vii/

// Time: O(N^2)
// Space: O(N^2)
class Solution {
public:
int stoneGameVII(vector<int>& A) {
int N = A.size();
vector<int> sum(N + 1);
for (int i = 0; i < N; ++i) sum[i + 1] = sum[i] + A[i];
vector<vector<int>> dp(N, vector<int>(N));
for (int len = 2; len <= N; ++len) {
for (int i = 0; i <= N - len; ++i) {
int j = i + len - 1;
dp[i][j] = max(sum[j + 1] - sum[i + 1] - dp[i + 1][j], sum[j] - sum[i] - dp[i][j - 1]);
}
}
return dp[0][N - 1];
}
};


Java

class Solution {
public int stoneGameVII(int[] stones) {
int length = stones.length;
int[] prefixSums = new int[length];
prefixSums[0] = stones[0];
for (int i = 1; i < length; i++)
prefixSums[i] = prefixSums[i - 1] + stones[i];
int[][] dp = new int[length][length];
for (int i = length - 2; i >= 0; i--) {
for (int j = i + 1; j < length; j++)
dp[i][j] = Math.max(getRangeSum(prefixSums, i + 1, j) - dp[i + 1][j], getRangeSum(prefixSums, i, j - 1) - dp[i][j - 1]);
}
return dp[0][length - 1];
}

public int getRangeSum(int[] prefixSums, int start, int end) {
return start == 0 ? prefixSums[end] : prefixSums[end] - prefixSums[start - 1];
}
}