lower_bound() in C++

Question

From reading from the Internet, I understand that The lower_bound() method in C++ is used to return an iterator pointing to the first element in the range [first, last) which has a value not less than value. This means that the function returns the index of the next smallest number just greater than that number.

So, for the given code below I understood that the output is 3. But, as there is repetition of 6. How can I get the index of last 6 using lower_bound(). I can implement my own binary_search() for that, but I want to know how to do it by lower_bound().

#include <iostream> 
#include <algorithm>
#include <vector> 

using namespace std; 

int main () 
{ 
    int array[] = {5,6,7,7,6,5,5,6}; 

    vector<int> v(array,array+8); // 5 6 7 7 6 5 5 6 

    sort (v.begin(), v.end()); // 5 5 5 6 6 6 7 7 

    vector<int>::iterator lower,upper; 
    lower = lower_bound (v.begin(), v.end(), 6); 
    upper = upper_bound (v.begin(), v.end(), 6); 

    cout << "lower_bound for 6 at position " << (lower- v.begin()) << '\n'; 
    return 0; 
} 

Answer 1

You can use reverse iterators into the vector, but then to fulfill the ordering requirement for std::lower_bound you need to inverse the comparison, so you need to use std::greater instead of the default std::less. This however also means that now you are not really looking for a lower bound, but for an upper bound with respect to that comparison function, so:

auto upper = std::upper_bound(v.rbegin(), v.rend(), 6, std::greater{});