Neighbour sums ring
Storage lockers are arranged in a ring, so the last locker sits next to the first. Given the parcel count in each locker (in ring order), return a list where position i holds the combined count of locker i's two neighbours: the one before and the one after, wrapping around the ends. With a single locker, both neighbours are that locker itself. Example: [1, 2, 3, 4] gives [6, 4, 6, 4].
Implement
ring_neighbor_sums(counts: list[int]) → list[int]Examples
in
[[1,2,3,4]]out[6,4,6,4]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 14 min
solution.py
InputExpectedGot
[[1,2,3,4]][6,4,6,4]not run yetsample