<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>

from typing import List

class Solution:
    def stoneGameII(self, piles: List[int]) -> int:
        n = len(piles)
        
        # Calculate the suffix sums to easily get the total stones from any position to the end
        suffix_sum = [0] * n
        suffix_sum[-1] = piles[-1]
        for i in range(n - 2, -1, -1):
            suffix_sum[i] = piles[i] + suffix_sum[i + 1]
        
        # Memoization table
        memo = {}
        
        def dp(i, M):
            # If we can take all remaining piles
            if i + 2 * M >= n:
                return suffix_sum[i]
            
            # If already computed, return the stored result
            if (i, M) in memo:
                return memo[(i, M)]
            
            # Try taking x piles where 1 <= x <= 2M
            max_stones = 0
            for x in range(1, 2 * M + 1):
                if i + x <= n:
                    # Calculate the stones Alice can get if Bob plays optimally
                    max_stones = max(max_stones, suffix_sum[i] - dp(i + x, max(M, x)))
            
            # Store the result in the memoization table
            memo[(i, M)] = max_stones
            return max_stones
        
        # Start the game from the first pile with M = 1
        return dp(0, 1)

# Example usage:
# solution = Solution()
# print(solution.stoneGameII([2, 7, 9, 4, 4]))  # Output: 10
# print(solution.stoneGameII([1, 2, 3, 4, 5, 100]))  # Output: 104
17800