clean, but the intital split/filter repetition seems a bit high in performance cost and very reduceable.

insane

20 lines? 🙂

very nooice

Pretty clean, love it

Or you could just use a.localeCompare(b) instead of trying to find the index. Misread that.

I'd say the best say is

[...'abcdefghijklmnopqrstuvwxyz']

How this test works? can you explain this?

To conclude: the best possible solution for this problem is a O(n) complexity (check my solution),
which is scaled out to O(2n) with this most "Clever"/"Best Practice".

The point is how s/he leverages pass by reference in his/her solution.

Thanks for your informative comment.

The right hand side of the || is only evaluated if the left hand side is zero (a falsy value). So the sorting prioritizes count (objects with higher count come first), and when both counts are equal, the next condition is used (alphabetical sorting, by src and then by char).

@voyxxx Did you mean 'fewer'?

This solution is Bonito (Beautiful)