Find all the elements which has no duplicate in multiple HashSet of Integers in Java

Question

In multiple HashSet of Integers I want to get all those elements, which has no duplicate. i.e. which came only once in union of all the HashSet. I am not able to conceptualize it programmatically.

As an example, consider set first contains {2,4,6,8,9}, second set contains {2,8,9} and third set contains {2,4,8,9}. In all these set, element 6 occurs only once.

How to find all the elements which has no duplicate in multiple HashSet of Integers in Java?

Answer 1

You could hold the set of elements that occur at least once and at least twice. It's a bit of manual looping but it's possible. This will work for any number of sets to difference and will not modify the input:

public static Set<E> unique(Set<? extends E>... sets){
   Set<E> once = new HashSet<E>();
   Set<E> twice = new HashSet<E>();

   for(Set<? extends E> set:sets){
      for(E el:set){
         if(once.contains(el)){
            twice.add(el);
         } else {
            once.add(el);
         }
      }
   }

   once.removeAll(twice);
   return once;
} 

Ideone: http://ideone.com/reGDBy

Example usage:

Set<Integer> set1, set2, set3;
...
Set<Integer> u = unique(set1, set2, set3);

Example of evaluation:

As an example, consider set first contains {2,4,6,8,9}, second set contains {2,8,9} and third set contains {2,4,8,9}. In all these set, element 6 occurs only once.

• After the first inner loop completes, once contains {2,4,6,8,9} and twice is empty.
• Adding the second set: 2, 8 and 9 are already in the once set, so they are added to the twice set.
• once is now {2,4,6,8,9}, twice is now {2,8,9}.
• From the third set: 2 is re-added to twice, 4 is added to twice, 8, 9 are re-added to twice.
• once is now {2,4,6,8,9} (union of all sets), twice is now {2,4,8,9} (elements that occur at least twice).
• remove twice from once. once is now {6}. Return once.