# Help me in solving CFRTEST problem

### My code

``````#include <iostream>
#include<bits/stdc++.h>
using namespace std;
#define ll long long

int main() {
ll int t;
cin>>t;

while(t--){
ll n;
cin>>n;

ll count = 0;
ll a[n];
for(int i=0; i<n; i++){
cin>>a[i];
}
sort(a , a+n);

for(int i=0; i<n; i++){
if(a[i] == a[i+1]){
count++;
}

}

if(count == 0){
cout<<n<<endl;
}
else if(count == (n-1)){
cout<<"1"<<endl;
}
else{
cout<<(count+ (n - count) - 1)<<endl;
}

}
return 0;
}

``````

Problem Link: CFRTEST Problem - CodeChef

@nayanshree28
you have understood the logic and solved 90% of the question but last part was not executed well.

just return n-count , that will be the answer . if 2 dates are same he can only save 1 friendship so increasing count by 1 will calculate how many friendships he cannot save.
so, the total friendships he can save is total friendships - no of he cannot save.

i have pasted my correct code below

hope this helps!!

include
#include<bits/stdc++.h>
using namespace std;

int main() {
int t;
cin>>t;
while(tâ€“)
{ int n,count=0;
cin>>n;
int a[n];
for(int i=0;i<n;i++)
{
cin>>a[i];
}
sort(a,a+n);
for(int i=0;i<n;i++)
{
if(a[i]==a[i+1])
count++;
}
int res=n-count;
std::cout << res << std::endl;

``````}
return 0;
``````

}