Question

Currently, I think my best option is to use std::set_intersection, and then check if the size of the smaller input is the same as the number of elements filled by set_intersection.

Is there a better solution?

Answer 1

Try this:

if (std::includes(set_one.begin(), set_one.end(),
                  set_two.begin(), set_two.end()))
{
// ...
}

About includes().

The includes() algorithm compares two
sorted sequences and returns true if
every element in the range [start2,
finish2) is contained in the range
[start1, finish1). It returns false
otherwise. includes() assumes that the
sequences are sorted using
operator<(), or using the predicate
comp.

runs in

At most ((finish1 – start1) + (finish2
– start2)) * 2 – 1 comparisons are performed.

Plus O(nlog(n)) for sorting vectors. You won’t get it any faster than that.