Alex and Lee continue their games with piles of stones.  There are a number of piles arranged in a row, and each pile has a positive integer number of stones piles[i].  The objective of the game is to end with the most stones. 

Alex and Lee take turns, with Alex starting first.  Initially, M = 1.

On each player’s turn, that player can take all the stones in the first X remaining piles, where 1 <= X <= 2M.  Then, we set M = max(M, X).

The game continues until all the stones have been taken.

Assuming Alex and Lee play optimally, return the maximum number of stones Alex can get.

Example 1:

Input: piles = [2,7,9,4,4]
Output: 10
Explanation:  If Alex takes one pile at the beginning, Lee takes two piles, then Alex takes 2 piles again. Alex can get 2 + 4 + 4 = 10 piles in total. If Alex takes two piles at the beginning, then Lee can take all three piles left. In this case, Alex get 2 + 7 = 9 piles in total. So we return 10 since it's larger. 

Constraints:

• 1 <= piles.length <= 100
• 1 <= piles[i] <= 10 ^ 4

The king of concise code reigns!

from functools import lru_cache

class Solution:
    def stoneGameII(self, piles: List[int]) -> int:
        
        N = len(piles)
        
        for i in range(N - 2, -1, -1):
            piles[i] += piles[i + 1]
            
        @lru_cache(None)
        def dp(i, m):
            if i + 2 * m >= N: 
                return piles[i]
            
            return piles[i] - min(dp(i + x, max(m, x)) for x in range(1, 2 * m + 1))
        
        return dp(0, 1)