Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
quest grind solution in Uncategorized category for When "k" is Enough! by Bifftastic
from typing import Iterable, Any
from collections import defaultdict
def remove_after_kth(items: list[Any], k: int) -> Iterable[Any]:
counter = defaultdict(int)
answer = []
for item in items:
counter[item] += 1
if counter[item] <= k:
answer.append(item)
return answer
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!")
Oct. 29, 2023
Comments: