Codes with frequency k
An inventory audit flags product codes by how often they were counted. Given the list of counted codes (in any order) and a target k (k >= 1), return all distinct codes that appear exactly k times, in ascending order. Return an empty list if none qualify. Example: [1, 2, 2, 3, 3, 3] with k = 2 gives [2].
Implement
codes_seen_k_times(codes: list[int], k: int) → list[int]Examples
in
[[1,2,2,3,3,3],2]out[2]What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
0:00 of about 12 min
solution.py
InputExpectedGot
[[1,2,2,3,3,3],2][2]not run yetsample