Day 14 Leet code series, today we will be picking the problem Maximum Number of Coins You Can Get (https://leetcode.com/problems/maximum-number-of-coins-you-can-get/).

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

class Solution {
public:
    int maxCoins(vector<int>& piles) {
        int ans=0;
        int count=0;
        int index=piles.size()-1;
        sort(piles.begin(), piles.end());
        while(count < piles.size()/3){
            ans+=piles[index-1];
            index-=2;
            count++;
        }
        return ans;
    }
};

Explaination:

1. First, I have sorted the given list
2. and taken 2nd Highest value from that array
3. because, We are allowed to take 2nd highest coin, So I have taken 2nd highest till n/3 times, and Decremented value of index by -2
4. Bob will always take remaining all less coins, So don’t care about Bob that which coins he will take.
5. So, we will take the values at -2 level rather than -3 level.