<p>Alice and Bob continue their games with piles of stones. There are a number of piles <strong>arranged in a row</strong>, and each pile has a positive integer number of stones <code>piles[i]</code>. The objective of the game is to end with the most stones.</p>

<p>Alice and Bob take turns, with Alice starting first.</p>

<p>On each player&#39;s turn, that player can take <strong>all the stones</strong> in the <strong>first</strong> <code>X</code> remaining piles, where <code>1 &lt;= X &lt;= 2M</code>. Then, we set <code>M = max(M, X)</code>. Initially, M = 1.</p>

<p>The game continues until all the stones have been taken.</p>

<p>Assuming Alice and Bob play optimally, return the maximum number of stones Alice can get.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">piles = [2,7,9,4,4]</span></p>

<p><strong>Output:</strong> <span class="example-io">10</span></p>

<p><strong>Explanation:</strong></p>

<ul>
	<li>If Alice takes one pile at the beginning, Bob takes two piles, then Alice takes 2 piles again. Alice can get <code>2 + 4 + 4 = 10</code> stones in total.</li>
	<li>If Alice takes two piles at the beginning, then Bob can take all three piles left. In this case, Alice get <code>2 + 7 = 9</code> stones in total.</li>
</ul>

<p>So we return 10 since it&#39;s larger.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">piles = [1,2,3,4,5,100]</span></p>

<p><strong>Output:</strong> <span class="example-io">104</span></p>
</div>

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

<ul>
	<li><code>1 &lt;= piles.length &lt;= 100</code></li>
	<li><code>1 &lt;= piles[i]&nbsp;&lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    vector<vector<vector<int>>> dp;
    int f(int p, int i, int m,vector<int>& piles){
        int n = piles.size(); 
        if(i>=n) return 0;

        if(dp[p][i][m]!=-1) return dp[p][i][m];

        int currSum = 0 , ans=0;
        int m1=0,m2=0;

        int mnsum=INT_MAX;
        int mxsum=INT_MIN;

        for(int j = i; j<min(n,i+2*m); j++){
            currSum+=piles[j];

            m1 = max(m,j-i+1);
            if(p==0)
                mxsum = max(mxsum,currSum+f(1,j+1,m1,piles));
            else
               mnsum = min(mnsum,f(0,j+1,m1,piles));
        }
        if(p==0) return dp[p][i][m] = mxsum;
        return dp[p][i][m]=mnsum;
    }
    int stoneGameII(vector<int>& piles){
        int n = piles.size();
        dp.resize(2,vector<vector<int>>(n,vector<int>(100,-1)));
        return f(0,0,1,piles);
    }
};
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