HackerEarth K Friends problem solution

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;
}