Max frequency stack
Design a stack that pops the most frequent element. Process ops: ['push', x] pushes x; ['pop'] removes and returns the most frequent element, breaking ties by returning the one pushed most recently among the maximally frequent. Return the list of popped values. pop is never called on an empty structure.
Implement
freq_stack(ops: list[list]) → list[int]Examples
in
[[["push",5],["push",7],["push",5],["push",7],["push",4],["push",5],["pop"],["pop"],["pop"],["pop"]]]out[5,7,5,4]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
[[["push",5],["push",7],["push",5],["push",7],["push",4],["push",5],["pop"],["pop"],["pop"],["pop"]]][5,7,5,4]not run yetsample