1561.MaximumNumberofCoinsYouCanGet.py

"""
    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 where piles[i] is the number of coins in 
    the ith pile.
    Return the maximum number of coins which you can have.

    Example:
    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.

    Example:
    Input: piles = [2,4,5]
    Output: 4

    Example:
    Input: piles = [9,8,7,6,5,1,2,3,4]
    Output: 18

    Constraints:
        - 3 <= piles.length <= 10^5
        - piles.length % 3 == 0
        - 1 <= piles[i] <= 10^4
"""
#Difficulty: Medium
#116 / 116 test cases passed.
#Runtime: 652 ms
#Memory Usage: 26.3 MB

#Runtime: 652 ms, faster than 60.00% of Python3 online submissions for Maximum Number of Coins You Can Get.
#Memory Usage: 26.3 MB, less than 40.00% of Python3 online submissions for Maximum Number of Coins You Can Get.

class Solution:
    def maxCoins(self, piles: List[int]) -> int:
        piles.sort()
        i = 0
        ans = 0
        length = len(piles)
        l = length // 3
        right = piles[l:]
        while i <= length - l - 1:
            ans += right[i]
            i += 2
        return ans