elegant way to remove all elements of a vector that are contained in another vector?

Question

While looking over some code I found loopy and algorithmically slow implementation of std::set_difference :

 for(int i = 0; i < a.size(); i++)
 {
  iter = std::find(b.begin(),b.end(),a[i]);
  if(iter != b.end())
  {
     b.erase(iter);
  }
 }

It can be easily replaced with sort(vectors are not sorted) + set_difference, but that requires allocation of new memory(see my recent Q Can output of set difference be stored in first input? why it cant be done "inplace").
So my solution would be something like:

sort(a.begin(), a.end());
for(size_t i = 0; i < b.size(); i++)
{
 if (binary_search(a.begin(), a.end(), b[i]))
 {
     swap(b[i], b[b.size()-1]); //remove current element by swapping with last
     b.pop_back();     // and removing new last by shrinking
 }
}

can it be done more elegantly?
elegant is subjective so within scope of this Q is defined as clearer code(ideally something from STL algorithms but I think it cant be done) but with no memory allocation and no increase in alg complexity.

Answer 1

Sort b so you can binary search it in order to reduce time complexity. Then use the erase-remove idiom in order to throw away all elements from a that are contained in b:

sort( begin(b), end(b) );
a.erase( remove_if( begin(a),end(a),
    [&](auto x){return binary_search(begin(b),end(b),x);}), end(a) );

Of course, you can still sacrifice time complexity for simplicity and reduce your code by removing the sort() and replacing binary_search() by find():

a.erase( remove_if( begin(a),end(a),
    [&](auto x){return find(begin(b),end(b),x)!=end(b);}), end(a) );

This is a matter of taste. In both cases you don't need heap allocations. By the way, I'm using lambda auto parameters which are C++14. Some compilers already implement that feature such as clang. If you don't have such a compiler, but only C++11 then replace auto by the element type of the container.

By the way, this code does not mention any types! You can write a template function so it works for all kind of types. The first variant requires random access iteration of b while the second piece of code does not require that.

Answer 2

This one does it in O(N+M), assuming both arrays are sorted.

  auto ib = std::begin(two);
  auto iter = std::remove_if (
       std::begin(one), std::end(one),
       [&ib](int x) -> bool {
                       while  (ib != std::end(two) && *ib < x) ++ib;
                       return (ib != std::end(two) && *ib == x);
                     });