c++ Is there STL algorithm to check if the range is strictly sorted?

Question

By "strictly" I mean "without equivalent elements".

is_sorted(v.begin(), v.end(), std::less<>()) 

doesn't meet my goal because it returns true for ranges like 1,2,2,4,5.

is_sorted(v.begin(), v.end(), std::less_equal<>()) 

would work according to the implementation given here, but unfortunately is_sorted requires Compare predicate to be strict ordering (Compare(a,a) must be false), and std::less_equal certainly isn't.

So should I write my own loop for this purpose?

Answer 1

Quoting the comment stream:

std::adjacent_find with std::greater_equal should do the trick - it will find the first element that is greater or equal than the next one; if no such element exists, the sequence is strictly increasing.

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

int main()
{
    std::vector<int> v1{0, 1, 2, 3, 40, 41};

    auto i2 = std::adjacent_find(v1.begin(), v1.end(), std::greater_equal<int>());
    if (i2 == v1.end()) {
        std::cout << "The entire vector is sorted in strictly ascending order\n";
    } else {
        std::cout << "The vector is not sorted\n";
    }
}

^{Example derived from http://en.cppreference.com/w/cpp/algorithm/adjacent_find}