<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:
    def stoneGameII(self, piles: List[int]) -> int:
        # M[i,m]: maximum number of stones alice can get for stone i to the end
        #with starting value m
        # res: M[0, 1]
        #for x in 1:2m 
        # M[i, m] = max(M[i, m], sum(piles[i:i+x]) + for y in 1:2*max(x, m) min(M[i+x+y,max(y,M)])
        #init: M[]
        suffix_sum = [0] * (len(piles) + 1)
        for i in range(len(piles) - 1, -1, -1):
            suffix_sum[i] = suffix_sum[i + 1] + piles[i]
        m = [[0] * (len(piles)+1) for _ in range(len(piles) + 1)]
        # Initialize the dp array.
        for i in range(len(piles)):
            m[i][len(piles)] = suffix_sum[i]
        for index in range(len(piles)-1, -1, -1): 
            for m_max in range(len(piles)-1, 0, -1):
                if index + m_max * 2 >= len(piles):
                    m[index][m_max] = suffix_sum[index]
                else:
                    for m_a in range(1, m_max*2 + 1):
                        m[index][m_max] = max(m[index][m_max], 
                                        suffix_sum[index] - m[index+m_a][max(m_max, m_a)])
        return m[0][1]

16600