If the input is not sorted, I think its a better idea to contruct a BST. Lets say you have first count query for the number (k) and now you have to evaluate it, construct BST at this point of time. If the next query is also of count, use the same BST. If the insertion query comes, using some space, store all the numbers. Till you get the next count query, keep storing numbers. As soon you get the next count query, make bst with all these elements you stored and delete the storage. This will come out to be O(nlogn) in total. This is less complex in understanding also, as compared to segment tree and BST.

You can use BIT and segment tree as said above, for understanding those, refer these:

Also, if the input is sorted, you can just use this segment code code(logarithm in complexity terms):

cin>>n;
for(int i=0;i<n;i++)
{
cin>>ele;
vec.push_back(ele);
}
sort(vec.begin(), vec.end());
it= lower_bound(vec.begin(), vec.end(),5); // returns iterator to first the entry in the vector which will not go before k, which here is 5.
int cnt= it-vec.begin(); // counting the number of elements
cout<<cnt<<endl;

yes i was talking about the same but you can also apply coordinate compression initially to reduce the efforts on applying binary search â€¦ because we just need to do comparision â€¦ so coordinate compression will not cause any harm to us â€¦

1.) Create segment tree every time a count query comes and you have done some insertion after the previous count query.
2.) Just make the segment tree once and use the concept of lazy propagation.

can you please elaborate how do i perform lazy updates in this context.With every insert query, the size of the array increases, so how do we operate on the old intervals?