you know we had to do it to em

this kind of code hints that we don't have friends. wakarimasu

This is the only one that works on infinite strings.

L M A O

Not gonna lie, it was quite difficult

really elegant.
for those not familiar with ViewPatterns, you can find related information here: https://gitlab.haskell.org/ghc/ghc/-/wikis/view-patterns.

in the above program, it is used to bind "y" with "show x"

very creative

Cool, but it lets me realize that the problem should not accept O(VE) solutions , it should accept only O(V + E * log(V)) instead

Vertices are removed one by one independently in each branch of computation. I'm not ready to estimate the complexity without much thinking, but I'm afraid it might be worse than exponential. This is just 101 edges: https://www.codewars.com/kumite/5c7993bcb79f2e46ed9f9f17

very clever solution!

I got a question about s == e, does it mean that one could go from 'a' to 'a' even 'a' is not in the graph?

Nope, It works in O(VE), where V - number of nodes, E - number of edges. Each time new node is visited all outgoing arcs are removed which requires traversing arc list, thus V times O(E) = O(VE)

Doesn't it have exponential time complexity?

hmmm, you solved this kata twice with the exact same solution?

You can recover the meaning from the test cases. It's difficult to describe them without giving away exactly the implementation.