Given a list of items, some of which may be duplicated, create and return a new Iterable that is otherwise the same as items, but only up to k occurrences of each element are kept, and all occurrences of that element after those first k are discarded. Note also the counterintuitive but still completely legitimate edge case of k == 0 that has a well defined answer of an empty list!
Input: A list and an integer.
Output: List or another Iterable (tuple, iterator, generator).
assert list(remove_after_kth([42, 42, 42, 42, 42, 42, 42], 3)) == [42, 42, 42] assert list(remove_after_kth([42, 42, 42, 99, 99, 17], 0)) ==  assert list(remove_after_kth([1, 1, 1, 2, 2, 2], 5)) == [1, 1, 1, 2, 2, 2]
This task is taken from the course CCPS 109 Computer Science I, as taught by Ilkka Kokkarinen.
CheckiO Extensions allow you to use local files to solve missions. More info in a blog post.
In order to install CheckiO client you'll need installed Python (version at least 3.8)
Install CheckiO Client first:
pip3 install checkio_client
Configure your tool
checkio --domain=py config --key=
Sync solutions into your local folder
(in beta testing) Launch local server so your browser can use it and sync solution between local file end extension on the fly. (doesn't work for safari)
checkio serv -d
Alternatevly, you can install Chrome extension or FF addon
checkio install-plugin --ff
checkio install-plugin --chromium