<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:
     int n;
    vector<vector<vector<int>>> dp;
     int solve(vector<int> &arr,int idx,int m,int alice){
        if(idx>=arr.size()){
            return 0;
        }
        if(dp[idx][m][alice]!=-1){
            return dp[idx][m][alice];
        }
        if(alice){
            int x = 1;
            int res = 0;
            int curSum = 0;
            while((x<=2*m) && (x+idx-1<=n-1)){
                  curSum+=arr[x+idx-1];
                  int ans = solve(arr,idx+x,max(m,x),false);
                  int newAns = curSum+ans;
                  res = max(res,newAns);
                  x++;
            }
            return dp[idx][m][alice]=res;
        }
        else{
            int x = 1;
            int res = INT_MAX;
            while((x<=2*m) && (x+idx-1<=n-1)){
                  int ans = solve(arr,idx+x,max(m,x),true);
                  int newAns = ans;
                  res = min(res,newAns);
                  x++;
            }
            return dp[idx][m][alice]=res;  
        }
        return -1;
     }
    int stoneGameII(vector<int>& piles) {
        n = piles.size();
        int m = 50;
        int alice = 2;
        dp.resize(n,vector<vector<int>>(m+1,vector<int>(alice,-1)));
        return solve(piles,0,1,true);
        
    }
};
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