<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+1,vector<vector<pair<int,int>>>(n+1,vector<pair<int,int>>(2,{0,0})));
        int total=0;
        for(int i=n-1;i>=0;i--)
        {
            total=stones[i];
            for(int j=i+1;j<n;j++)
            {
                total+=stones[j];
                for(int k=0;k<2;k++)
                {
                    if(i>=j)
                    {
                       dp[i][j][k]={0,0};
                       continue;
                    }
                pair<int,int>a=dp[i+1][j][k^1];
                pair<int,int>b=dp[i][j-1][k^1];
                if(k==0)
                {
                    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{
                    if(a.first-(total-stones[i]+a.second)<=b.first-(total-stones[j]+b.second))
                    {
                        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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