<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>

// Time:  O(n*(logn)^2)
// Space: O(nlogn)

class Solution {
public:
    int stoneGameII(vector<int>& piles) {
        for (int i = piles.size() - 2; i >= 0; --i) {
            piles[i] += piles[i + 1];
        }
        unordered_map<int, unordered_map<int, int>> lookup;
        return dp(piles, &lookup, 0, 1);
    }

private:
    int dp(const vector<int>& piles,
           unordered_map<int, unordered_map<int, int>> *lookup,
           int i, int m) {
        if (i + 2 * m >= piles.size()) {
            return piles[i];
        }
        if (!lookup->count(i) || !(*lookup)[i].count(m)) {
            int min_dp = numeric_limits<int>::max();
            for (int x = 1; x <= 2 * m; ++x) {
                min_dp = min(min_dp, dp(piles, lookup, i + x, max(m, x)));
            }
            (*lookup)[i][m] = piles[i] - min_dp;
        }
        return (*lookup)[i][m];
    }
};

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