What is the fastest way to compare two sets in Java?

by 
firstSet.equals(secondSet)

It really depends on what you want to do in the comparison logic… ie what happens if you find an element in one set not in the other? Your method has a void return type so I assume you’ll do the necessary work in this method.

More fine-grained control if you need it:

if (!firstSet.containsAll(secondSet)) {
  // do something if needs be
}
if (!secondSet.containsAll(firstSet)) {
  // do something if needs be
}

If you need to get the elements that are in one set and not the other.

EDIT: set.removeAll(otherSet) returns a boolean, not a set. To use removeAll(), you’ll have to copy the set then use it.

Set one = new HashSet<>(firstSet);
Set two = new HashSet<>(secondSet);
one.removeAll(secondSet);
two.removeAll(firstSet);

If the contents of one and two are both empty, then you know that the two sets were equal. If not, then you’ve got the elements that made the sets unequal.

You mentioned that the number of records might be high. If the underlying implementation is a HashSet then the fetching of each record is done in O(1) time, so you can’t really get much better than that. TreeSet is O(log n).

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