What is an efficient and elegant way to add a single element to an immutable set?

Question

I have an immutable set (cast as a Set) that potentially contains many elements. I need a Collection that contains the elements from that set plu

Answer 1

I'm experiencing cognitive dissonance when I read "immutable" and "add to" in the same sentence. You can add a new element to the end of a mutable copy of the immutable values, but you can't modify the immutable set. I don't know of anything elegant.

Answer 2

You might consider Sets.union(). Construction would be faster, but use slower.

public static <T> Set<T> setWith(Set<T> old, T item) {
  return Sets.union(old, Collections.singleton(item);
}

(com.google.common.collect.Sets & java.util.Collections)

Answer 3

You have three options.

• Use a mutable set.
• Check the element isn't already present, if not create a copy of the set and add an element.
• Create a wrapper set which includes the previous set and the element.

Sometimes a BitSet is a better choice than Set<Integer> depending on the distribution of your values.

Answer 4

Using Java 8 you can also use streams for that effect

Stream.concat(oldSet.stream(),
              Stream.of(singleElement))
      .collect(toSet())

Answer 5

Not sure about performance, but you can use Guava's ImmutableSet.Builder:

import com.google.common.collect.ImmutableSet

// ...
Set<Integer> newSet = new ImmutableSet.Builder<Integer>()
                                .addAll(oldSet)
                                .add(3)
                                .build();

Of course you can also write yourself a helper method for that:

public static <T> Set<T> setWith(Set<T> old, T item) {
  return new ImmutableSet.Builder<T>().addAll(old).add(item).build();
}

// ...
Set<Integer> newSet = setWith(oldSet, 3);

Answer 6

If the Set is immutable, I don't see any way to do it other than copy the Set, and then add your new element. Remember, copying a set is as easy as passing the base set to the constructor function when creating the new set.