Fixed

added

Approved

one is enough

Fixed.

Fixed in latest fork

rejected. see comment on your fork.

I think it's clever because I was really trying to figure out a way to solve it without relying on lambda
but couldn't quite figure it out because if two items have the same key, their original order will be preserved. However, by reversing the words, I inadvertently changed the order for words with the same last character. Surely this example here is not best practice but a unique approach I appreciate.

Complexity here I think is:
O(n×m), where n is the length of the alphabet and m is the number of words in listx

approved.

.

approved

I'm very glad you think so.

5 seems fair