<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:
        CACHE = {}

        def max_stones(pile_no=-1, M=1):
            cache_key = (pile_no, M)
            if cache_key in CACHE:
                return CACHE[cache_key]

            max_extra_piles = 2*M
            best_turn_total_stones = 0
            best_oth_turn_total_stones = 0
            turn_stones = 0

            for x in range(1, max_extra_piles+1):
                new_pile_no = pile_no + x
                if new_pile_no >= len(piles):
                    break

                turn_stones += piles[new_pile_no]
                future_stones_res = max_stones(
                    new_pile_no, max(M,x)
                )

                future_stones = future_stones_res[1]
                future_oth_stones = future_stones_res[0]
                turn_total_stones = future_stones + turn_stones

                if turn_total_stones > best_turn_total_stones:
                    best_turn_total_stones = turn_total_stones
                    best_oth_turn_total_stones = future_oth_stones
            
            result =  [best_turn_total_stones, best_oth_turn_total_stones]
            # max stones current player can get,
            # stones that other player would get in optimal play
            CACHE[cache_key] = result
            return result

        return max_stones()[0]

                
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