<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 {
    int solve(int turn, int index, int m, vector<int>& piles, vector<vector<vector<int>>>& dp)
    {
        //Base Case
        if(index >= piles.size())
        {
            return 0;
        }

        if(dp[turn][index][m] != -1)
        {
            return dp[turn][index][m];
        }

        int res = turn == 0 ? 0 : INT_MAX;
        int sum = 0;

        int t = min((int)piles.size()-index, 2 * m);

        for(int x=1;x <= t; x++)
        {
            sum = sum + piles[index + x - 1];
            if(turn == 0){
               res = max(res, sum + solve(!turn, index+x,max(m,x),piles,dp)); 
            }
            else
            {
                res = min(res, solve(!turn, index+x,max(m,x),piles,dp));
            }
        }

        return dp[turn][index][m] = res;

    }
public:
    int stoneGameII(vector<int>& piles) {
        int n = piles.size();
        vector<vector<vector<int>>> dp(2, vector<vector<int>>(n,vector<int>(n+1,-1)));
        return solve(0,0,1,piles, dp);
    }
};
15321