This is still O(array1.length * array2.length) - just as yours.
Replacing loops with streams is not taking away complexity.

Yes it is. The list after the filtering is not gauranteed to be sorted.

See my reply john777

Cool so to explain the filter function a little. The filter function takes in a function that returns TRUE or FALSE. The inner lambda expression (as a whole) returns a boolean VALUE (Arrays.stream(array2).anyMatch(s -> s.contains(str)).

For example I can write
boolean matchedResult = memberNames.stream().anyMatch((s) -> s.startsWith("filtered-strings-have-this-text"));

Notice again that matchedResults is of a boolean type, although anyMatch takes in a function as an argument, it returns a definate answer. (true or false)

The magic happens because this value (str) in the filter function satisfies the condition f(str) -> True or False.

Took me a little while to wrap my head around that ... coming frm a 1.7 world.

Guys, remember that this solution is better than all other solutions because it's complexity is on a smaller order. Other solutions AFAIK are on an order of O(n^2) including mine.