Is my approach fine or there is any better approach

#1

#include
#include
using namespace std;
bool compare(pair<int,int>p1,pair<int,int>p2)
{
return p1.second<p2.second;
}
int main()
{
int t,n,n1,n2;
int mx,count;
cin>>t;
while(t–)
{
mx=0;
cin>>n;
pair<int,int>p[n];
for(int i=0;i<n;i++)
{
cin>>n1>>n2;
p[i]=make_pair(n1,n2);
}
sort(p,p+n,compare);
for(int i=0;i<n;i++){
count=1;
for(int j=i+1;j<n;j++)
{
if(p[i].second<=p[j].first)
{
i=j;
count++;
}
}
if(count>mx)
mx=count;
}
cout<<mx<<endl;
}
return 0;
}

#2

for(int i=0;i<n;i++){
count=1;
for(int j=i+1;j<n;j++)
{
if(p[i].second<=p[j].first)
{
i=j;
count++;
}

You don’t need to do it in a loop, a single traversal is enough,i.e., you need to pick greedily.
Hint you will always pick the first activity in your solution.

#3

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