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First solution in Clear category for Remove All After by lmaiaux
from collections.abc import Iterable
def remove_all_after(items: list[int], border: int) -> Iterable[int]:
return items[:items.index(border)+1] if border in set(items) else items
print("Example:")
print(list(remove_all_after([1, 2, 3, 4, 5], 3)))
# These "asserts" are used for self-checking
assert list(remove_all_after([1, 2, 3, 4, 5], 3)) == [1, 2, 3]
assert list(remove_all_after([1, 1, 2, 2, 3, 3], 2)) == [1, 1, 2]
assert list(remove_all_after([1, 1, 2, 4, 2, 3, 4], 2)) == [1, 1, 2]
assert list(remove_all_after([1, 1, 5, 6, 7], 2)) == [1, 1, 5, 6, 7]
assert list(remove_all_after([], 0)) == []
assert list(remove_all_after([7, 7, 7, 7, 7, 7, 7, 7, 7], 7)) == [7]
print("The mission is done! Click 'Check Solution' to earn rewards!")
June 14, 2023
