How to check if Set B has at least one element that is not in Set A

Question

So I have two sets: A and B. I need to check if set B contains anything that is not in the set A. There are maybe intersections, so I cannot just check if set A contains set B. I can obviously do this:

for (String string : setA) {
  if (!setB.contains(string) {
    break;
  }
}

or using the Guava library:

Sets.intersection(setA, setB).containsAll(setB); // returns false if there are elements outside.

But is there any way that would perform better or may be just cleaner or more elegant? Thanks.

Answer 1

“B contains an element not in A” is the exact opposite of “A contains all elements of B”, therefore, the already existing method containsAll is sufficient to answer that question.

if(!setA.containsAll(setB)) {
    System.out.println("setB contains an element not in setA");
}

You may shortcut using setB.size()>setA.size() || !setA.containsAll(setB), but this requires that the sets agree on the definition of equality, e.g. if one set is a SortedSet using String.CASE_INSENSITIVE_ORDER as comparator and the other is a HashSet, this won’t work (but the definition of the correct outcome is tricky with such combinations anyway).

If setB is really large, you might get a benefit from using a parallel stream like

if(!setB.parallelStream().allMatch(setA::contains)) {
    System.out.println("setB contains an element not in setA");
}

but this is rather rare.

Answer 2

Merge all elements into another set and compare the total elements:

Set ab = new Set(a);
ab.addAll(b);

if (ab.size() != b.size()) break; // that means `a` had some element that was not in b