Lower and Upper Bound in case of Decreasing Non-ascending vector

Question

For an Increasing/Ascending array, I grasped the notion of Lower and Upper discovered.

Lower Bound: iterator pointing to the first element in the [first, last] range >= Value

Upper Bound: iterator pointing to the first element in the [first, last] range > Value

Below is my Decreasing/Non-ascending vector code, which is causing me problems:

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

using namespace std;

int main()
{
    vector<int> vec = {45,40,35,12,6,3};

    auto itr3 = lower_bound(vec.begin(),vec.end(),40);
    auto itr4 = upper_bound(vec.begin(),vec.end(),40);

    if(itr3 == vec.end() && itr4 == vec.end())
        cout<<"Lower & Upper Bound not Found\n";
    else
    {
        cout <<"lower Bound of 40 :"<<*itr3<<endl;
        cout <<"Upper Bound of 40 :"<<*itr4<<endl;
    }

    return 0;
}

The Output is:

Lower & Upper Bound not Found.

But as mentioned above the output should be something like :

lower Bound of 40 :40
Upper Bound of 40 :45

Please assist me in understanding the lower and upper bound behaviour in the situation of decreasing/non-ascending vectors.

Answer 1

Both std::lower bound and std::upper bound must have an increasing (non-decreasing) order as their objective.

By giving a comparator as the 4th parameter of the functions, you may modify the meaning of "growing."

To work with descending vectors, use std::greater.

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

using namespace std;

int main()
{
    vector<int> vec = {45,40,35,12,6,3};

    auto itr3 = lower_bound(vec.begin(),vec.end(),40,std::greater<int>());
    auto itr4 = upper_bound(vec.begin(),vec.end(),40,std::greater<int>());

    if(itr3 == vec.end() && itr4 == vec.end())
        cout<<"Lower & Upper Bound not Found\n";
    else
    {
        cout <<"lower Bound of 40 :"<<*itr3<<endl;
        cout <<"Upper Bound of 40 :"<<*itr4<<endl;
    }

    return 0;
}