<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:
        # Method 1: Dynamic Programming (Top-Down) (Memoization). Time O(n³) Space O(n²)
        n = len(piles)
        memo = [[0] * (n + 1) for _ in range(n)]
        suffixSum = [0] * (n + 1)

        for i in range(n):
            suffixSum[i + 1] = suffixSum[i] + piles[i]

        @cache
        def dfs(index, move):
            if 2 * move >= n - index:
                return suffixSum[n] - suffixSum[index]
            if memo[index][move]:
                return memo[index][move]
            opponent = 0
            for x in range(1, 2 * move + 1):
                opponent = max(
                    opponent, suffixSum[n] - suffixSum[index] - dfs(index + x, max(move, x))
                )
            memo[index][move] = opponent
            return opponent

        return dfs(0, 1)

    def _stoneGameII(self, piles: List[int]) -> int:
        # Method 2: Dynamic Programming (Bottom-Up) (Tabulation). Time O(n³) Space O(n²)
        n = len(piles)
        dp = [[0] * (n + 1) for _ in range(n + 1)]
        suffixSum = [0] * (n + 1)

        for i in range(n - 1, -1, -1):
            suffixSum[i] = suffixSum[i + 1] + piles[i]

        for i in range(n - 1, -1, -1):
            for move in range(1, n + 1):
                for x in range(1, 2 * move + 1):
                    if i + x <= n:
                        dp[i][move] = max(
                            dp[i][move], suffixSum[i] - dp[i + x][max(move, x)]
                        )

        return dp[0][1]
19300