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one liner solution in Creative category for When "k" is Enough! by Sim0000
from typing import Iterable
def remove_after_kth(items: list, k: int) -> Iterable:
return [e for i, e in enumerate(items) if items[:i+1].count(e) <= k]
print("Example:")
print(list(remove_after_kth([42, 42, 42, 42, 42, 42, 42], 3)))
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!")
Oct. 21, 2022
