# Leetcode - Maximum Number of Coins You Can Get Solution

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 `i`

pile.^{th}

Return the maximum number of coins which 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 <= 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))
```