There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows:
- In each step, you will choose any 3 piles of coins (not necessarily consecutive).
- Of your choice, Alice will pick the pile with the maximum number of coins.
- You will pick the next pile with maximum number of coins.
- Your friend Bob will pick the last pile.
- Repeat until there are no more piles of coins.
Given an array of integers
piles[i] is the number of coins in the
Return the maximum number of coins which you can have.
Input: piles = [2,4,1,2,7,8] Output: 9 Explanation: Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with 7 coins and Bob the last one. Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with 2 coins and Bob the last one. The maximum number of coins which you can have are: 7 + 2 = 9. On the other hand if we choose this arrangement (1, 2, 8), (2, 4, 7) you only get 2 + 4 = 6 coins which is not optimal.
Input: piles = [2,4,5] Output: 4
Input: piles = [9,8,7,6,5,1,2,3,4] Output: 18
3 <= piles.length <= 10^5
piles.length % 3 == 0
1 <= piles[i] <= 10^4
Solution in Python
class Solution: def maxCoins(self, piles: List[int]) -> int: n = len(piles) sorted_piles = sorted(piles)[::-1] return sum(sorted_piles[i] for i in range(1, n//3*2+1,2))