# 1561. Maximum Number of Coins You Can Get

## 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.

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

## Solutions

Greedy.

• class Solution {

public int maxCoins(int[] piles) {
Arrays.sort(piles);
int ans = 0;
for (int i = piles.length - 2; i >= piles.length / 3; i -= 2) {
ans += piles[i];
}
return ans;
}
}

• class Solution {
public:
int maxCoins(vector<int>& piles) {
sort(piles.begin(), piles.end());
int ans = 0;
for (int i = piles.size() - 2; i >= (int) piles.size() / 3; i -= 2) ans += piles[i];
return ans;
}
};

• class Solution:
def maxCoins(self, piles: List[int]) -> int:
piles.sort()
return sum(piles[-2 : len(piles) // 3 - 1 : -2])


• func maxCoins(piles []int) int {
sort.Ints(piles)
ans, n := 0, len(piles)
for i := n - 2; i >= n/3; i -= 2 {
ans += piles[i]
}
return ans
}

• function maxCoins(piles: number[]): number {
piles.sort((a, b) => a - b);
const n = piles.length;
let ans = 0;
for (let i = 1; i <= Math.floor(n / 3); i++) {
ans += piles[n - 2 * i];
}
return ans;
}


• impl Solution {
pub fn max_coins(mut piles: Vec<i32>) -> i32 {
piles.sort();
let n = piles.len();
let mut ans = 0;
for i in 1..=n / 3 {
ans += piles[n - 2 * i];
}
ans
}
}