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.
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.
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)