Elements over n/3
Given an integer array a, return all elements that appear strictly more than floor(n/3) times, where n is the array length. Return the answers as a list sorted in ascending order. There can be at most two such elements; return an empty list if none qualify.
Implement
majority_n3(a: list[int]) → list[int]Examples
in
[[3,2,3]]out[3]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 30 min
solution.py
InputExpectedGot
[[3,2,3]][3]not run yetsample