<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:
    unordered_map<int, int> cache;
    int helper(bool is_alice, int i, int M, vector<int>& piles){
        int key = i*1000+M;
        if (not is_alice) key*=-1;
        if (cache.count(key)) return cache[key];
        int N = piles.size();
        int left = N-i+1;
        if (left<=2*M){
            if (not is_alice) return 0;
            int amt = 0;
            for (i; i<N; i++) amt+= piles[i];
            cache[key] = amt;
            return amt;
        }
        int amt = 0, res =0;
        if (not is_alice) res = INT_MAX;
        int limit = min(i+2*M,N);
        for (int j = i; j<limit; j++){
            int taken = j-i+1;
            left = helper(!is_alice, j+1, max(taken,M), piles);
            if (is_alice){
                amt += piles[j];
                res = max(res,left+amt);
                continue;
            }
            res = min(left,res);
        }
        cache[key] = res;
        return res;
    }
    int stoneGameII(vector<int>& piles) {
        // f('a', i, M)
        // left = N-i+1;
        // if left<=2*M then take everything
        // then dp
        return helper(true, 0, 1, piles);

    }
};
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