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First solution in Clear category for When "k" is Enough! by Pavellver
from typing import Iterable, Any
def remove_after_kth(items: list[Any], k: int) -> Iterable[Any]:
lst = []
for i in items:
if lst.count(i) < k:
lst.append(i)
return lst
print("Example:")
print(list(remove_after_kth([42, 42, 42, 42, 42, 42, 42], 3)))
# These "asserts" are used for self-checking
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]
print("The mission is done! Click 'Check Solution' to earn rewards!")
March 8, 2023
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