<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<pair<int,int>>>> dp;
    int n;

    pair<int,int> solve(vector<int>&a, int chance, int m, int i){
        if(i>=n) return {0,0};
        
        auto &p = dp[chance%2][m][i];
        if(p.first != -1 && p.second != -1) return p;

        int ans1= 0, ans2 =0 , sum=0;
        if(chance%2 == 0) ans2=INT_MAX;
        else ans1 = INT_MAX;

        for(int x=0; x<2*m && i+x < n; x++){
            sum += a[i+x];
            auto p = solve(a, chance+1, max(x+1,m), i+x+1);

            if(chance%2 == 0){
                ans1 = max(ans1, sum + p.first);
                ans2 = min(ans2 , p.second);
            } else{
                ans1 = min(ans1 , p.first);
                ans2 = max(ans2 , sum + p.second);
            }

        }

        return p = {ans1 , ans2};
    }

    int stoneGameII(vector<int>& piles) {
        n = piles.size();
        dp = vector<vector<vector<pair<int,int>>>>(2, 
                vector<vector<pair<int,int>>> (2*n+1 , 
                    vector<pair<int,int>>(n+1 ,{-1,-1})));

        auto p = solve(piles, 0, 1, 0);
        return p.first;
    }
};
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