检查unordered_set是否包含其他unordered_set中的所有元素-C ++

Question

I'm brand new to C++ and have been asked to convert a Java program to C++. I'm trying to write a method to check that all elements in an unordered_set exist in another unordered_set. I found the example below using hash_set but hash_set is deprecated and it is recommended to use unordered_set now.

// returns true if one contains all elements in two
bool SpecSet::containsAll(hash_set<Species*> one, hash_set<Species*> two) {
   sort(one.begin(), one.end());
   sort(two.begin(), two.end());
   return includes(one.begin(), one.end(), two.begin(), two.end());
}

So I need a way to do this using unordered_set. Sort does not work on unordered sets and lookup speed is important, so I don't want to use an ordered set.

bool SpecSet::containsAll(unordered_set<Species*> one, unordered_set<Species*> two) {

   return ?;
}

I'd really appreciate some help with an approach to doing this efficiently.

EDIT: I guess this will work. It seems there is no more efficient way but to loop over all in two.

bool SpecSet::containsAll(unordered_set<Species*> one, unordered_set<Species*> two) {
   if(two.size() > one.size())
   {
      return false;
   }

   for(Species *species : two)
   {
      if(one.find(species) == one.end())
      {
         return false;
      }
   }
   return true;
}

Answer 1

Disclaimer: This is not the most efficient approach. It is an attempt at solution that would be as generic and flexible as std::includes while supporting unordered iterator ranges. It is not limited to std::unordered_set and should work for any other container, eg, std::vector or std::list .

---

As it was pointed out std::includes requires the input ranges to be sorted. At this moment unordered ranges are not supported in the standard library.

Looking at possible implementations of std::includes a version for unordered ranges can be implemented. For example like so:

template<class InputIt1, class InputIt2>
bool includes_unordered(
    InputIt1 first1, InputIt1 last1,
    InputIt2 first2, InputIt2 last2)
{
    for (; first2 != last2; ++first2)
    {
        InputIt1 it1;
        for (it1 = first1; it1 != last1; ++it1)
        {
            if(*first2 == *it1)
                break;
        }
        if (it1 == last1)
            return false;
    }
    return true;
}

Note: containers' size-comparison optimization is not performed to support containers of non-unique objects. But if needed it can be done using std::distance .

And here's a version taking an equivalence operator:

template<class InputIt1, class InputIt2, class Equivalence>
bool includes_unordered(
    InputIt1 first1, InputIt1 last1,
    InputIt2 first2, InputIt2 last2,
    Equivalence equiv)
{
    for (; first2 != last2; ++first2)
    {
        InputIt1 it1;
        for (it1 = first1; it1 != last1; ++it1)
        {
            if(equiv(*first2, *it1))
                break;
        }
        if (it1 == last1)
            return false;
    }
    return true;
}

Small live-example

Then includes_unordered can be used in the same way as std::includes would.

Answer 2

With unsorted collections, there's no faster algorithm than to iterate over the smaller collection while testing that each element is a member of the larger. This will naturally scale as O( n ), where n is the size of the putative subset, since we perform an O(1) find operation n times.

---

Here's some demonstration code, with tests:

#include <unordered_set>

template <typename T>
bool is_subset_of(const std::unordered_set<T>& a, const std::unordered_set<T>& b)
{
    // return true if all members of a are also in b
    if (a.size() > b.size())
        return false;

    auto const not_found = b.end();
    for (auto const& element: a)
        if (b.find(element) == not_found)
            return false;

    return true;
}

int main()
{
    const std::unordered_set<int> empty{ };
    const std::unordered_set<int> small{ 1, 2, 3 };
    const std::unordered_set<int> large{ 0, 1, 2, 3, 4 };
    const std::unordered_set<int> other{ 0, 1, 2, 3, 9 };

    return 0
        +  is_subset_of(small, empty) // small ⊄ ∅
        + !is_subset_of(empty, small) // ∅ ⊂ small
        +  is_subset_of(large, small) // large ⊄ small
        + !is_subset_of(small, large) // small ⊂ large
        +  is_subset_of(large, other) // large ⊄ other
        +  is_subset_of(other, large) // other ⊄ large
        + !is_subset_of(empty, empty) // ∅ ⊂ ∅
        + !is_subset_of(large, large) // x ⊂ x, ∀x
        ;
}

---

An equivalent, using standard algorithm instead of writing an explicit loop:

#include <algorithm>
#include <unordered_set>

template <typename T>
bool is_subset_of(const std::unordered_set<T>& a, const std::unordered_set<T>& b)
{
    // return true if all members of a are also in b
    auto const is_in_b = [&b](auto const& x){ return b.find(x) != b.end(); };

    return a.size() <= b.size() && std::all_of(a.begin(), a.end(), is_in_b);
}

(obviously using the same main() for tests)

---

Note that we pass the sets by reference , not by value, as you've indicated that the sets are too large for you to copy and sort them.