<p>There are <code>3n</code> piles of coins of varying size, you and your friends will take piles of coins as follows:</p>

<ul>
	<li>In each step, you will choose <strong>any </strong><code>3</code> piles of coins (not necessarily consecutive).</li>
	<li>Of your choice, Alice will pick the pile with the maximum number of coins.</li>
	<li>You will pick the next pile with the maximum number of coins.</li>
	<li>Your friend Bob will pick the last pile.</li>
	<li>Repeat until there are no more piles of coins.</li>
</ul>

<p>Given an array of integers <code>piles</code> where <code>piles[i]</code> is the number of coins in the <code>i<sup>th</sup></code> pile.</p>

<p>Return the maximum number of coins that you can have.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> piles = [2,4,1,2,7,8]
<strong>Output:</strong> 9
<strong>Explanation: </strong>Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with <strong>7</strong> coins and Bob the last one.
Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with <strong>2</strong> 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, <strong>2</strong>, 8), (2, <strong>4</strong>, 7) you only get 2 + 4 = 6 coins which is not optimal.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> piles = [2,4,5]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> piles = [9,8,7,6,5,1,2,3,4]
<strong>Output:</strong> 18
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= piles.length &lt;= 10<sup>5</sup></code></li>
	<li><code>piles.length % 3 == 0</code></li>
	<li><code>1 &lt;= piles[i] &lt;= 10<sup>4</sup></code></li>
</ul>


#pragma GCC optimize("O3")
class Solution {
public:
    int maxCoins(vector<int>& piles) {
        int freq[10000+1]={0};
        int M=0;
        for(int i=0;i<piles.size();i++)
        {
            freq[piles[i]]++;
            M=max(M,piles[i]);
        }
        int count=0,x;
        int alice=0;
        int ans=0;
        
        for(x=M;count<piles.size()/3;x--)
        {
            if(freq[x])
            {   
                int f1=freq[x]/2;//half will be mine
                if(freq[x]%2)//for odd coins
                {
                    alice=alice?0:1;
                    f1+=!alice;
                }
                ans+=f1*x;
                count+=f1;
            }
        }
        ans-=(x+1)*(count-piles.size()/3);
        return ans;

    }
};
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