Common elements between two lists | Kuchup interview prep

The 30 second answer

I put the second list into a HashSet, then filter the first list with contains and distinct. That is linear expected time instead of nested scans. I mention equals and hashCode again if the element type is a custom class.

Set<T> setB = new HashSet<>(b);
List<T> common = a.stream()
    .filter(setB::contains)
    .distinct()
    .toList();

O(n + m) expected time instead of O(n · m) from calling contains on a list. [Opinion — standard complexity reasoning]

Sources:

Stream.filter / distinct

Sources (0)

See inline citations in the answer above.