In this **HackerEarth K Friends problem solution** Monk has N friends. They are invited to his birthday party. Each friend has a satisfying factor which is equal to the number of gifts which they are expecting. Monk wants to satisfy at-least K friends but he is unaware of their satisfying factors. So Monk starts distribution of gifts. As soon as a friend is satisfied he won't take more gifts.

Monk will follow a distribution strategy so as to minimize the number of gifts needed to satisfy atleast K of his friends. Find the minimum number of gifts which Monk should carry with himself in the worst case.

## HackerEarth K Friends problem solution.

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
ll n,k,t;
vector<ll>v;
int main()
{
freopen("inp10.txt","r",stdin);
freopen("out10.txt","w",stdout);
ll i,j,ans=0,cur;
cin>>t;
while(t--)
{
cin>>n;
v.clear();
ans=cur=0;
for(i=1;i<=n;i++)
{
cin>>j;
v.push_back(j);
}
sort(v.begin(),v.end());
cin>>k;
i=0;
while(k--)
{
ans+=((n-i)*(v[i]-cur));
cur=v[i];
i++;
}
cout<<ans<<"\n";
}
return 0;
}

### Second solution

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int n,a[100005];
cin>>n;
for(int i=0;i<n;i++)
cin>>a[i];
int k;
cin>>k;
sort(a,a+n);
ll ans=0;
for(int i=0;i<k;i++)
{
ans+=(ll)a[i];
}
for(int i=k;i<n;i++)
ans+=(ll)a[k-1];
cout<<ans<<"\n";
}
return 0;
}

Thursday, January 13, 2022