<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 __init__(self):
        # A map of whose turn it is, which index in the game we are at, 
        # and how many steps the player can take 
        # -> to how many points the players will have from that turn.
        self.memo = {}
        self.piles = None

    def nextTurn(self, player: bool, start: int, limit: int) -> int:
        memoKey =  (player, start, limit)
        if memoKey in self.memo:
            return self.memo[memoKey]
        if start + limit >= len(self.piles):
            if player:
                self.memo[memoKey] = (0,sum(self.piles[start:]))
            else:
                self.memo[memoKey] = (sum(self.piles[start:]),0)
            return self.memo[memoKey]
        roundScore = 0
        maxScore = (0,0)
        for i in range(limit):
            roundScore += self.piles[start+i]
            p0Score, p1Score = self.nextTurn(not player, start+i+1, max(limit, (i+1)*2))
            if player:
                candScore = roundScore + p1Score
                if candScore > maxScore[1]:
                    maxScore = (p0Score, candScore)
            else:
                candScore = roundScore + p0Score
                if candScore > maxScore[0]:
                    maxScore = (candScore, p1Score)
        
        self.memo[memoKey] = maxScore
        return self.memo[memoKey]


    def stoneGameII(self, piles: List[int]) -> int:
        self.piles = piles
        result = self.nextTurn(False,0,2)
        return result[0]
        
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