# PROBLEM LINK:

**Setter:** Nisarg Thakkar

**Tester:** Malhar Patel

**Editorialist:** Nisarg Thakkar

# DIFFICULTY:

Easy-medium

# PREREQUISITES:

Nothing

# PROBLEM:

The idea behind the problem boils down to following : You are given detail of T teams and you need to find the team having maximum potential(including all dancers of the team).

# EXPLANATION

If you closely observe the problem, if you first call the dancer having maximum potential in the team on the stage then, though his potential would be counted 0 but due to that next two dancers will be benefited, as they will get the potential of first dancer as per the question. Thus in this manner the overall potential of the group will be maximum.

The for any given array the answer will be sum of N terms in a manner 0(1st dancer) + max(input_arr) + second max of array + second max of array and so on.

Thus in this manner the potential will be maximized.

# SOLUTIONS:

## CPP Solution

#include <bits/stdc++.h>

using namespace std;

#define ll long long int

#define pb push_back

#define llu unsigned long long int

#define fo(i,n) for(int i=0;i<n;i++)

#define eb emplace_back

#define M 1000000007

#define vi vector

#define vlli vector

#define pi pair<int,int>

#define mp make_pair

#define mapi map<int,int>

ll max(ll x,ll y)

{

if(x>y)

return x;

return y;

}

ll min(ll x,ll y)

{

if(x<y)

return x;

return y;

}

void yes()

{

cout << “YES” << endl;

}

void no()

{

cout << “NO” << endl;

}

bool isPrime(ll n)

{

if (n <= 1)

return false;

if (n <= 3)

return true;

if (n % 2 == 0 || n % 3 == 0)

return false;

```
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
```

}

int main()

{

ios::sync_with_stdio(false);

cin.tie(0);

cout.tie(0);

//cin.ignore(numeric_limits::max(),’\n’);

//const unsigned int M=1000000007;

ll test,ans=0;

cin >> test;

//test=1;

while(test–)

{

ll n,cnt=0;

cin >> n;

vector a(n);

fo(i,n) cin>>a[i];

sort(a.begin(),a.end());

reverse(a.begin(),a.end());

int i=0,x=1;

while(x<=n && i<n)

{

if(i==0) cnt+=a[i], x++;

else

{

cnt+=a[i];

x++;

if(x==n) break;

cnt+=a[i];

x++;

if(x==n) break;

}

i++;

if(x==n) break;

}

ans = max(ans,cnt);

// cout<<cnt<<endl;

}

cout << ans << endl;

return 0;

}

Feel free to Share your approach, if you want to. (even if its same ) . Suggestions are welcomed as always had been.