<p>Alice and Bob play a game with piles of stones. There are an <strong>even</strong> number of piles arranged in a row, and each pile has a <strong>positive</strong> integer number of stones <code>piles[i]</code>.</p>

<p>The objective of the game is to end with the most stones. The <strong>total</strong> number of stones across all the piles is <strong>odd</strong>, so there are no ties.</p>

<p>Alice and Bob take turns, with <strong>Alice starting first</strong>. Each turn, a player takes the entire pile of stones either from the <strong>beginning</strong> or from the <strong>end</strong> of the row. This continues until there are no more piles left, at which point the person with the <strong>most stones wins</strong>.</p>

<p>Assuming Alice and Bob play optimally, return <code>true</code><em> if Alice wins the game, or </em><code>false</code><em> if Bob wins</em>.</p>

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

<pre>
<strong>Input:</strong> piles = [5,3,4,5]
<strong>Output:</strong> true
<strong>Explanation:</strong> 
Alice starts first, and can only take the first 5 or the last 5.
Say she takes the first 5, so that the row becomes [3, 4, 5].
If Bob takes 3, then the board is [4, 5], and Alice takes 5 to win with 10 points.
If Bob takes the last 5, then the board is [3, 4], and Alice takes 4 to win with 9 points.
This demonstrated that taking the first 5 was a winning move for Alice, so we return true.
</pre>

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

<pre>
<strong>Input:</strong> piles = [3,7,2,3]
<strong>Output:</strong> true
</pre>

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

<ul>
	<li><code>2 &lt;= piles.length &lt;= 500</code></li>
	<li><code>piles.length</code> is <strong>even</strong>.</li>
	<li><code>1 &lt;= piles[i] &lt;= 500</code></li>
	<li><code>sum(piles[i])</code> is <strong>odd</strong>.</li>
</ul>

class Solution {
public:
    
    bool stoneGame(vector<int>& p) {
        int n = p.size();
        vector<vector<pair<int,int>>> adp(n, vector<pair<int,int>>(n,make_pair(-1,-1))), bdp(n, vector<pair<int,int>>(n,make_pair(-1,-1)));
        vector<vector<int>> afdp(n, vector<int>(n,-1));
        int res =  opgame(p,afdp,0, p.size()-1, true);   
       int sum = 0;
       for (int i = 0 ; i< n ; i++)
       {
        sum+=p[i];
       }
       
       for (int i = 0 ; i< n ; i++)
       {
            for (int j = 0 ; j< n ; j++)
            {
                cout<<afdp[i][j]<<" ";        
            }
            cout<<endl;
       }
        


        return afdp[0][p.size()-1]>sum-afdp[0][p.size()-1];
    }

   int opgame(vector<int>& p,vector<vector<int>>& afdp,int st , int et, bool a)
    {
        if (st>et || st<0 || et>=p.size())
        {
            return 0;
        }

        if (st==et)
        {
            if (a)
                return p[st];
            else
                return 0;
        }

        if (a)
        {
            if (afdp[st][et] == -1)
            {
                int r1 = opgame(p,afdp, st+1, et, !a);
                int r2 = opgame(p,afdp, st, et-1, !a);
                 afdp[st][et] = r1+p[st] > r2+ p[et]?r1+p[st] : r2+ p[et];
            }
            return afdp[st][et] ;
        }
        else
        {
             if (afdp[st][et] == -1)
            {
                int r1 = opgame(p,afdp, st+1, et, !a);
               int r2 = opgame(p,afdp, st, et-1, !a);
               afdp[st][et] = r1 < r2?r1: r2;
            }
            return afdp[st][et] ;
        }
    }
};
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