<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:
        prefix_sum = [0 for i in range(len(piles)+1)]
        for i in range(len(piles)):
            prefix_sum[i+1] = prefix_sum[i] + piles[i]
        @lru_cache(maxsize=None)
        def minmax(i, m, player):
            if (i + 2*m) >= len(piles):
                return prefix_sum[-1] - prefix_sum[i] if player else 0
            if player:
                stones = float("-inf")
                for j in range(1, (2*m)+1):
                    take = prefix_sum[i+j] - prefix_sum[i]
                    stones = max(stones, take + minmax( i+j, max(m, j), not player))
                return stones
            else:
                stones = float("inf")
                for j in range(1, (2*m)+1):
                    stones = min(stones, minmax( i+j, max(m, j), not player))
                return stones
        return minmax(0, 1, True)

   


'''
[2,7,9,4,4]
[0, 2, 9, 18, 22, 26] 
'''
20400