<p>Alice and Bob take turns playing a game, with <strong>Alice starting first</strong>.</p>

<p>There are <code>n</code> stones arranged in a row. On each player&#39;s turn, they can <strong>remove</strong> either the leftmost stone or the rightmost stone from the row and receive points equal to the <strong>sum</strong> of the remaining stones&#39; values in the row. The winner is the one with the higher score when there are no stones left to remove.</p>

<p>Bob found that he will always lose this game (poor Bob, he always loses), so he decided to <strong>minimize the score&#39;s difference</strong>. Alice&#39;s goal is to <strong>maximize the difference</strong> in the score.</p>

<p>Given an array of integers <code>stones</code> where <code>stones[i]</code> represents the value of the <code>i<sup>th</sup></code> stone <strong>from the left</strong>, return <em>the <strong>difference</strong> in Alice and Bob&#39;s score if they both play <strong>optimally</strong>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> stones = [5,3,1,4,2]
<strong>Output:</strong> 6
<strong>Explanation:</strong> 
- 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> stones = [7,90,5,1,100,10,10,2]
<strong>Output:</strong> 122</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == stones.length</code></li>
	<li><code>2 &lt;= n &lt;= 1000</code></li>
	<li><code>1 &lt;= stones[i] &lt;= 1000</code></li>
</ul>

class Solution {
public:
    int stoneGameVII(vector<int>& stones) {
        int n = stones.size();
        vector<vector<vector<pair<int,int>>>> dp(n, vector<vector<pair<int,int>>>(n, vector<pair<int,int>>(2, {0, 0})));
        vector<int> prefixSum(n + 1, 0);
        
        // Compute prefix sum
        for (int i = 0; i < n; ++i) {
            prefixSum[i + 1] = prefixSum[i] + stones[i];
        }

        for (int length = 2; length <= n; ++length) {
            for (int i = 0; i + length - 1 < n; ++i) {
                int j = i + length - 1;
                int total = prefixSum[j + 1] - prefixSum[i];  // Total sum from i to j
                
                for (int k = 0; k < 2; ++k) {
                    pair<int, int> a = dp[i + 1][j][k ^ 1];
                    pair<int, int> b = dp[i][j - 1][k ^ 1];
                    
                    if (k == 0) {  // Player 1's turn
                        if (total - stones[i] + a.first - a.second >= total - stones[j] + b.first - b.second) {
                            dp[i][j][k] = a;
                            dp[i][j][k].first += total - stones[i];
                        } else {
                            dp[i][j][k] = b;
                            dp[i][j][k].first += total - stones[j];
                        }
                    } else {  // Player 2's turn
                        if (abs(total - stones[i] + a.second - a.first) >= abs(total - stones[j] + b.second - b.first)) {
                            dp[i][j][k] = a;
                            dp[i][j][k].second += total - stones[i];
                        } else {
                            dp[i][j][k] = b;
                            dp[i][j][k].second += total - stones[j];
                        }
                    }
                }
            }
        }

        return dp[0][n-1][0].first - dp[0][n-1][0].second;
    }
};

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