I don't know

Please specify the language when raising issues. I'm assuming you're referring to the C translation.

[...] I would've never guessed that it should be of the same length of the input string.

No, the size of the allocated buffer is irrelevant as long as it's big enough. What's important is to set the value of nb_vowels appropriately (refer to the solution setup).

Not a kata issue. Next time, please raise a Question when you're unsure.

I reduced the input size to max 300 elements.

They were removed because the kata was decided to not be a performance kata. It would be quite inconsistent if some languages allowed slow solutions while others enforced performance. I really don't know if Rust can trivially handle 10_000 element arrays even with very slow solutions, but the intent is that performance shouldn't be on solvers minds at all (or at least, it shouldnt need to be).

In my original Python translation, hard tests could go up to 10_000 elements, but Kacarott removed those tests in his fork. Rust is much faster than Python, I tried to make roughly equivalent tests in both languages, though I have no real idea about what a naive solution would give.

0..100_000 is the length of the array, not the values inside it. The values are random signed 32 bits integers (rng.gen::<i32>()).

I've forked it with the changes you wanted

Essentially, yes. You have to sort and you have to allocate at some point.

Because lst is a slice of u8s and slices are non-owning by definition, calling lst.iter() results in an iterator of references to u8s (Iterator<Item = &u8>).

The return type of the delete_nth function is Vec<u8>, which can be built through .collect() only from an iterator over owned values (Iterator<Item = u8>).

The iterator methods which allow such a conversion (from Iterator<Item = &u8> to Iterator<Item = u8>) are .cloned() and .copied() (either one works in this case).

Strings are immutable in Python, so, what you do inside your function with its arguments is up to you.

In Python N is as big as 1000000000. Note this:

Hint: You're not meant to calculate the factorial. Find another way to find the number of zeros.