My first Solution was in fact to compare every element of b with the squares of a. XD After reading this comment, I thought about a better solution and wrote a second solution with a hashtable. I think, that is the most efficient way. Well, maybe not my implementation. ;)
Thanks anyway for the inspiration.

If you compare b to a without sorting it would be O(n^2), with sorting it's O(2nlog n + n) = O(n log n) which is better for large arrays.

@Sourdough, what's about this?))

Thanks!

pat pat~

See solutions
Crying out of misery

Literally everyone of us, with some kata, at some point of time