<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:
        n = len(piles)
        total_s = sum(piles)

       # memo = [[None] * ((n + 3) // 4 + 1) for _ in range(n)]

        def dfs(pos, max_take):
            if n == pos:
                return 0
            
            if memo[pos][max_take] is not None:
                return memo[pos][max_take]
            
            max_num = -sys.maxsize
            # max_num = 0
            s = 0
            for i in range(pos, min(n, pos + 2 * max_take)):
                s += piles[i]
                next_take = max(max_take, i - pos + 1)
                next_take = min(next_take, (n + 3) // 4)
                max_num = max(max_num, s - dfs(i + 1, next_take))
            
            memo[pos][max_take] = max_num
            return max_num
        
        # ret = dfs(0, 1)
        # x - y = ret and x + y = s =>  2x= (s + ret)=>x = (s + ret) // 2
        # return (s + ret) // 2
        size = (n + 3) // 4
        dp = [[0] * (size + 1) for _ in range(n + 1)]
        for pos in range(n - 1, -1 , -1):
            limit = (n - pos + 1) // 2
            for max_take in range(1, min(limit + 1, (n + 3) // 4 + 1)):
                s, max_num = 0, -sys.maxsize
                for i in range(pos, min(n, pos + 2 * max_take)):
                    s += piles[i]
                    next_take = max(max_take, i - pos + 1)
                    next_take = min(next_take, size)
                    # next_take = min(next_take, (n - (i + 1) + 1) // 2)
                    next_limit = min(next_take, (n - (i + 1) + 1) // 2)
                    max_num = max(max_num, s - dp[i + 1][next_limit])
                
                dp[pos][max_take] = max_num
            
            for max_taken in range(limit + 1, size + 1):
                dp[pos][max_taken] = dp[pos][limit]

        return (total_s + dp[0][1]) // 2

            

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