<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:
        sld={}
        def helper(stones,limit):
            if (len(stones),limit) in sld:
                return sld[(len(stones),limit)][0],sld[(len(stones),limit)][1]
            if limit>=len(stones):
                player1score = sum(stones)
                player2score = 0

            player1score=0
            player2score=0
            sumsofar = 0
            nl = limit
            for i in range(min(limit,len(stones))):
                sumsofar = sumsofar+stones[i]
                if (i+1)*2>limit:
                    nl = (i+1)*2
                rscore1,rscore2=helper(stones[i+1:],nl)
                if rscore2+sumsofar>player1score:
                    player1score = rscore2+sumsofar
                    player2score=rscore1
            sld[(len(stones),limit)]=[player1score,player2score]
            return player1score,player2score



        p1,p2 = helper(piles,2)
        return p1
19500