<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 stoneGameII(vector<int>& piles) {
        vector<vector<vector<int>>> memo(2, vector<vector<int>>(100, vector<int>(100, -1)));
        return maxStones(true, 0, 1, piles, memo);
    }

    int maxStones(bool a, int i, int m, vector<int>& piles, vector<vector<vector<int>>>& memo){
        if(i >= piles.size()) return 0;
        if(memo[a][i][m] != -1) return memo[a][i][m];
        if(a){
            int total = 0;
            int currentMax = 0;
            for(int x = 1; x <= 2*m && (i + x) <= piles.size(); x++){
                total += piles[i - 1 + x];
                currentMax = max(total + maxStones(false, i + x, max(x, m), piles, memo), currentMax);            
            }
            cout<<"ms :: a="<<a<<" i="<<i<<" m="<<m<<" res="<<currentMax<<endl;
            memo[a][i][m] = currentMax;
        } else {
            if(i >= piles.size() - 1) {
                memo[a][i][m] = 0;
                return 0;
            }
            int currentMin = INT_MAX;
            for(int x = 1; x <= 2*m && (i + x) <= piles.size(); x++){
                currentMin = min(maxStones(true, i + x, max(x, m), piles, memo), currentMin);
            }
            cout<<"ms :: a="<<a<<" i="<<i<<" m="<<m<<" res="<<currentMin<<endl;
            memo[a][i][m] = currentMin;
        }
        return memo[a][i][m];
    }

    int max(int a, int b){
        if(a > b) return a;
        else return b;
    }

    int min(int a, int b){
        if(a < b) return a;
        else return b;
    }
};
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