maximum-number-of-coins-you-can-get.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 the maximum number of coins.
# 	Your friend Bob will pick the last pile.
# 	Repeat until there are no more piles of coins.
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# Given an array of integers piles where piles[i] is the number of coins in the ith pile.
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# Return the maximum number of coins that you can have.
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#  
# Example 1:
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# 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.
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# Example 2:
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# Input: piles = [2,4,5]
# Output: 4
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# Example 3:
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# Input: piles = [9,8,7,6,5,1,2,3,4]
# Output: 18
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#
#  
# Constraints:
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# 	3 <= piles.length <= 105
# 	piles.length % 3 == 0
# 	1 <= piles[i] <= 104
#
#


class Solution:
    def maxCoins(self, piles: List[int]) -> int:
        piles.sort(reverse=True)
        res = 0
        d = deque(piles)
        while d:
            d.popleft()
            d.pop()
            res += d.popleft()
        return res