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First solution in Clear category for When "k" is Enough! by suic
from typing import Iterable
def remove_after_kth(items: list, k: int) -> Iterable:
seen: dict[int, int] = {}
for item in items:
if seen.get(item, 0) >= k:
continue
if item not in seen:
seen[item] = 0
seen[item] += 1
yield item
Oct. 21, 2022
