<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 play(vector<int>& piles, int M, int n, int remaining, unordered_map<int, unordered_map<int, int>>& results)
    {
        if (n >= piles.size())
            return 0;

        if (results.find(n) != results.end() && results.find(n)->second.find(M) != results.find(n)->second.end())
            return results.find(n)->second.find(M)->second;

        int acc = 0;
        int result = numeric_limits<int>::min();

        if (2 * M >= piles.size())
            result = remaining;
        {
            for (int X = 1; X <= 2 * M; ++X)
            {
                if (n - 1 + X == piles.size())
                    break;

                acc += piles[n - 1 + X];
                int interResult = remaining - play(piles, max(M, X), n + X, remaining - acc, results);

                if (interResult > result)
                    result = interResult;
            }
        }

        if (results.find(n) == results.end())
            results.insert(pair<int, unordered_map<int, int>>(n, unordered_map<int, int>()));

        results.find(n)->second.insert(pair<int, int>(M, result));
        
        return result;
    }

    int stoneGameII(vector<int>& piles) 
    {
        int remaining = 0;
        for (int x : piles)
            remaining += x;

        unordered_map<int, unordered_map<int, int>> results;
        return play(piles, 1, 0, remaining, results);
    }
};
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