1561. Maximum Number of Coins You Can Get

URL: LeetCode Problem

Problem Description

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 the 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 that you can have.

Examples:

• Example 1: 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.

Constraints:

• 3 <= piles.length <= 105
• piles.length % 3 == 0
• 1 <= piles[i] <= 104

Python3 Solution

class Solution:
    def maxCoins(self, piles: List[int]) -> int:
        piles.sort()
        q = deque(piles)
        total = 0
        while len(q) > 0:
            q.pop()
            total += q.pop()
            q.popleft()
        return total