deworde - 2 months ago 8x

C++ 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

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.

Source (Stackoverflow)

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