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.
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.
Example 2:
Input: piles = [2,4,5]
Output: 4
Example 3:
Input: piles = [9,8,7,6,5,1,2,3,4]
Output: 18
Constraints:
-
3 <= piles.length <= 105
-
piles.length % 3 == 0
-
1 <= piles[i] <= 104
SOLUTION:
class Solution:
def maxCoins(self, piles: List[int]) -> int:
n = len(piles)
piles.sort()
score = 0
i = 0
j = n
while j - i >= 3:
score += piles[j - 2]
i += 1
j -= 2
return score
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