Top k frequent items
Given a stream of string items, return the k most frequent items. Order the result by frequency descending; break ties by the item string ascending (lexicographic). Return a list of [item, count] pairs of length min(k, number of distinct items). If k <= 0 return an empty list. The stream may be empty.
Implement
top_k_frequent(items: list[str], k: int) → listExamples
in
[["a","b","a","c","b","a"],2]out[["a",3],["b",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 35 min
solution.py
InputExpectedGot
[["a","b","a","c","b","a"],2][["a",3],["b",2]]not run yetsample