Count elements lower than a given value in a std::set

The correct way to do a lower bound search is with std::set's own lower_bound function, which is specially designed to work with this sorted, associative, non-random-access container.

So, instead of this:

std::lower_bound( mySet.begin(), mySet.end(), 2 );

use this:

mySet.lower_bound(2);

This is logarithmic in the size of the container, which is much better than a std::count_if approach (which doesn't know about the sortyness of the comparator, and therefore must visit all the nodes and is thus linear).

However, you then must also use std::distance from the beginning to the lower bound, which is not only linear but also necessarily "slow" in practice (due to the non-random-access).

Nathan's solution seems optimal given that you do not want to simply find the lower bound, but find its distance from the "start" of the container.

Looks like operator- is not valid for a std::set::iterator. Why?

Indeed, an implementation of std::set::iterator::operator-() can't exist in constant complexity since the elements are not contiguous in memory.

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Then, how can you easily (without calling std::previous or std::next until bound is reached...this would not be efficient) find the position of a given iterator in the container?

You can't, std::set::iterator is not a RandomAccessIterator. See std::distance() documentation:

Complexity

Linear.

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If you can't, then what alterantive can I use to find the index of a given element...?

I'd suggest to count your elements without having to compute an iterator distance: std::count_if() can help us:

#include <iostream>
#include <algorithm>
#include <set>

int main()
{
    std::set<int> mySet;
    mySet.insert( 1 );
    mySet.insert( 2 );
    mySet.insert( 3 );
    mySet.insert( 4 );

    const std::size_t lower_than_three = std::count_if(
         std::begin(mySet)
        , std::end(mySet)
        , [](int elem){ return elem < 3; } );
    std::cout << lower_than_three << std::endl;    
}

Demo

Since std::set::iterator is a BidirectionalIterator we cannot subtract from it unless we use the decrement operator. What we can do though is just walk the set and count the iterations until we reach a number bigger than what we are looking for.

std::set<int> mySet;
// fill values
int counter = 0;
for (auto it = mySet.begin(), *it < some_value && it != mySet.end(); ++it)
{
    if (e < some_value)
        counter++;
}

This is a worst mySet.size() iterations which is as fast as you can get it when dealing with a BidirectionalIterator.

Also note that std::lower_bound does not have O(log N) complexity since we are not using a RandomAccessIterator. When using a non-RandomAccessIterator it has linear complexity.