Why not do away with the if entirely and write n.left and queue.append(n.left)? That might be obfuscation though, depending on who you ask.

Me either, was expecting a clever mathematical trick ¯_(ツ)_/¯

Should be ok now. Thanks for reporting!

Yeah man. Pattern matching is a great feature.

Very clever way of checking if the list is size one. You could also destructure x and y as (x:xs) and (y:ys), in which case you can check whether xs or ys is null.

Thank you!

Thank you!

The reference solution checks for equality, so I'm pretty sure the tests are sound.

The thing is, most shifts other than the correct shift will not produce words. Therefore if your solution can get most of the words, it will succeed. To me, this is normal behavior.

Why so many random tests?

Thank you both! My math background is a little spotty, sometimes simple stuff flies right over my head...!

I've rewritten the solution again.

Note that the answer in your example is actually 22 : you forgot to count a 'v'.