<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 solve(int i, int M, int isAlice, vector<vector<vector<int>>> &dp, vector<int> &piles){
        if(i==piles.size()-1 && isAlice){
            return piles[i];
        }
        else if(i==piles.size()-1){
            return 0;
        }
        else if(i>=piles.size()){
            return 0;
        }
        if(dp[i][M][isAlice]!=-1){
            return dp[i][M][isAlice];
        }
        int j = i;
        int sum = 0;
        int mini = INT_MAX;
        while(isAlice && j < i+(2*M) && j<piles.size()){
            sum += piles[j];
            dp[i][M][isAlice] = max(dp[i][M][isAlice], sum + solve(j+1, max(j-i+1, M), 0, dp, piles));
            j++;
        }
        if(!isAlice){
            
            while( j < i+(2*M) && j<piles.size()){
                
                mini = min(mini, solve(j+1, max(M,j-i+1) , 1, dp, piles));
                j++;
            }
            dp[i][M][isAlice] = mini;
        }
        return dp[i][M][isAlice];

         }
    int stoneGameII(vector<int>& piles) {
        int n = piles.size();
        vector<vector<vector<int>>> dp(n, vector<vector<int>>(n, vector<int> (2, -1)));
        return solve(0, 1, 1, dp, piles);

    }
};
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