Here's an example of input whose expected output does not follow the "one vowel per line" rule:
"oGweQFMoJNk KnJLHOtimUtRTu GirCkKZBLIzXtOfTfNvfI h"

If the test conformed to the description, the last line of expected output would be "fTfNvfI h". But the expected output instead puts a newline after the I and leaves h on the last line by itself.

In testing and modifying my code to handle cases like this, I found that it was the space following the I that was triggering the newline. If that space were not there, the last line of the expected output would be "fTfNvfIh". This is not explained in the description.

You should really not mark issues as "Resolved" until they have been resolved. I think you first need to reach a clear understanding in your own mind about the rules for the expected output. Then these rules will be easier to express accurately in the description.

that still leaves 2) 3) 4) 6) unaddressed

6) on my list is such a case:

input:    'a5 '
expected: 'a-\n5 '
but got:  'a5 '

and it also doesn't hold when there are \n's in the input

furthermore the description contradicts that by saying that there should be a \n after each vowel

if you say that there should be exactly one vowel on each line, then one needs to be able to split it into lines and find a vowel in each line. that is not the case.

OK, but I saw at least one case where the test did require a new line after the final vowel if there was a space after that final vowel. That should be explained.

Danke gleichfalls!

The reason that the sentence exists is because I want "Ho-\nme-\s" for "Homes" to be wrong and "Ho-\nmes" not
Also the reason why this is not the case is because my solution might have counter other characters as letters. So string.ascii_letters is used instead of string.printable for the random tests

Schönen Tag noch!

I used string.printable instead of just ascii letters. Just using ascii letters for the random tests now ✅

Oh yeah and the 5th case is totally a problem on my solution. Will try to fix it

The last vowel not needing to be broken apart is the reason why the "The final result should have exactly one vowel per line." sentence is in the description. Since if there would be a line break after the last vowel, there would be no vowel in every line anymore

though unclear which

That's the reason why I included Merriam Webster's definition of when y count as a vowel:

The word has no other vowel

which I interpreted as "the word not having a, e, i, o, or u"

So for szyldyk (which I have replaced to szyldy due to @hobovsky pointing out that szyldyk doesn't actually exists), both ys count as vowels

My solution might have counted other characters as letters. So I replaced string.printable in the random tests with just ascii letters

✅ Replaced string.printable with just ascii letters

My solution might count other characters as letter, which is probably why cases like were handled that way

The description states: "The final result should have exactly one vowel per line." This is clearly not the case in the expected results for many of the tests.

Tip: Read the notes in the "Modulo check confirmation example" very carefully. There is a bit of important information there that occurs nowhere else in the description.

Thanks!

Approved at 7 kyu based on the consensus, but some will have to think a little harder than usual for a 7 kyu. I especially liked the extra information provided as a reward for solving it.

I couldn't get it out of my head

good that it didn't let you down after all