LFU cache
Simulate an LFU (least-frequently-used) cache of the given capacity over a list of operations. Each op is ['put', key, value] or ['get', key]. 'get' returns the value or -1 if absent and counts as a use; 'put' inserts/updates and counts as a use. When at capacity and inserting a new key, evict the least-frequently-used key; break ties by least-recently-used among the minimum-frequency keys. If capacity is 0, every put is a no-op. Return the list of results for every 'get' op, in order.
Implement
lfu_cache(capacity: int, ops: list[list]) → list[int]Examples
in
[2,[["put",1,1],["put",2,2],["get",1],["put",3,3],["get",2],["get",3],["put",4,4],["get",1],["get",3],["get",4]]]out[1,-1,3,-1,3,4]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 40 min
solution.py
InputExpectedGot
[2,[["put",1,1],["put",2,2],["get",1],["put",3,3],["get",2],["get",3],["put",4,4],["get",1],["get",3],["get",4]]][1,-1,3,-1,3,4]not run yetsample