<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:
    global dp, p, ans
    def rec(self,isAliceTurn, x, m):
        global dp, p, ans
        # print(isAliceTurn, x, m)
        if x==len(p):
            return 0
        if dp[isAliceTurn][x][m]!=0:
            return dp[isAliceTurn][x][m]
        for i in range(x,min(len(p), 2*m+x)):
            if isAliceTurn:
                
                dp[1][x][m]=max( sum(p[x:])-self.rec(0,i+1,max(i+1-x,m)) , dp[1][x][m])
            else:
                dp[0][x][m] = max(sum(p[x:])-self.rec(1, i+1, max(i+1-x,m )),dp[0][x][m])
        # print(isAliceTurn, x, m, 'ans-> ', dp[isAliceTurn][x][m])
        return dp[isAliceTurn][x][m]
            


    def stoneGameII(self, piles: List[int]) -> int:
        global dp,p,ans
        ans=0
        p=piles[::]
        dp=[[[0 for i in range(201)] for j in range(101)] for k in range(2)]
        return self.rec(1, 0, 1)
        
        
17500