<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>>(piles.size(), vector<int>(2 * piles.size(), -1)));
        return dfs(true, 0, 1, piles, memo);
    }

    int dfs(bool isA, int i, int M, vector<int>& piles, vector<vector<vector<int>>>& memo) {
        if (i == piles.size()) {
            return 0;
        }
        if (memo[isA][i][M] != -1) {
            return memo[isA][i][M];
        }

        int response;
        if (isA) {
            response = 0;
        }
        else {
            response = INT_MAX;
        }
        int sum = 0;

        for (int X = 1; X <= 2 * M; X++) {
            if (i + X > piles.size()) {
                break;
            }
            sum += piles[i + X - 1];
            if (isA) {
                response = max(response, sum + dfs(!isA, i + X, max(M, X), piles, memo));
            }
            else {
                response = min(response, dfs(!isA, i + X, max(M, X), piles, memo));
            }
        }
        memo[isA][i][M] = response;
        return response;
    }
};
19528