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?
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.
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