Find employees without badges
Facilities needs to chase employees who never picked up a badge. `employees` rows look like "id,name" (ids and names are distinct); `badge_ids` is a list of employee ids that already have a badge (it may contain ids not in the employee list). Return the names of employees whose id is NOT in `badge_ids`, sorted alphabetically. Example: employees ["e1,ava", "e2,raj", "e3,mia"] with badge_ids ["e2"] gives ["ava", "mia"].
Implement
missing_badges(employees: list[str], badge_ids: list[str]) → list[str]Examples
in
[["e1,ava","e2,raj","e3,mia"],["e2"]]out["ava","mia"]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 11 min
solution.py
InputExpectedGot
[["e1,ava","e2,raj","e3,mia"],["e2"]]["ava","mia"]not run yetsample