<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 {
    unordered_map<int,unordered_map<int,int>>dp;
    int memo(vector<int>& piles,int m,int ind,int n){
        if(ind>=n)return 0;
        int rem=n-ind+1;
        m=min(m,(rem/2)+rem%2);

        if(dp.count(ind)==1 && dp[ind].count(m)==1)return dp[ind][m];
        int maxi = INT_MIN;
        int sum = 0;

        for(int x = 1;x<=2*m ;x++){
            if(ind+x-1>=n)break;
            sum += piles[ind+x-1];
            int newM = max(x,m);
            int mini=INT_MAX;
            for(int oppX = 1;oppX<=2*newM;oppX++){
                mini=min(mini,memo(piles,max(newM,oppX),ind+x+oppX,n));
            }
            maxi=max(maxi,sum+mini);
        }
        return dp[ind][m]=maxi;

    }
public:
    int stoneGameII(vector<int>& piles) {
        return memo(piles,1,0,piles.size());
    }
};
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