<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 mem[100][100][2];

    int rec(vector<int>& arr, int pos, int m, int isAlice) {
        if (mem[pos][m][isAlice]) {
            return mem[pos][m][isAlice];
        }

        int n = arr.size();
        bool isLast = pos + 2*m >= n;
        int sm = 0;
        vector<int> v;

        for(int i=1; i<=2*m && i+pos-1<n; ++i) {
            if (isAlice) {
                sm+=arr[pos+i-1];
            }
            if (!isLast) {
                v.push_back(rec(arr, pos+i, max(m, i), !isAlice) + sm);
            }
        }

        int res;
        if (isLast || v.size() == 0) {
            res = sm;
        } else {
            res = v[0];
            for(auto item : v) {
                res = isAlice ? max(res, item) : min(res, item);
            }
        }

        mem[pos][m][isAlice] = res;

        return res;
    }

    int stoneGameII(vector<int>& piles) {
        return rec(piles, 0, 1, 1);
    }
};
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