Drone Delivery on a Hub Ring — Minimum Travel

OA problem. `m` hubs sit on a ring (1-indexed); a drone starts at Hub 1 and must visit `requestedHubs` in order, each leg taking the shorter of the clockwise / counter-clockwise path measured by per-edge transition times. Return the minimum total delivery time. The classic traps are the tail-to-head wrap-around and the `m == 2` ring edge case. Requirements m hubs are arranged on a ring, 1-indexed; the ring closes (the last hub connects back to the first). Input: the ordered list of requestedHubs to visit, and a list of transition times for the edges between adjacent hubs (the cost of leaving each hub for its neighbor). The drone starts at Hub 1 and visits the requested hubs in the given order. For each leg it may travel clockwise or counter-clockwise and pays the cheaper of the two directional path costs. Return the minimum total travel time to complete every requested visit. Notes Precompute prefix sums of the edge transition times around the ring so any clockwise arc cost is O(1); the counter-clockwise cost for the same pair is total_ring_cost - clockwise_cost. The per-leg answer is the min of the two. The tail-to-head step is the most-reported bug: moving from the last hub back toward Hub 1 must use the same wrap-around prefix-sum arithmetic, not a special case. Several candidates lose points exactly here. The m == 2 ring is a degenerate case (two parallel edges between the same pair); confirm how its two directions are defined before coding. Mind whether edge costs are uniform or per-edge — when transition times differ per edge, you cannot shortcut with hop-count distance; you must compare the two summed arc costs. Recent HackerRank OA wording is long and domain-heavy; budget time to strip it down to ring + requested destinations + adjacent-edge distances before coding the prefix-sum helper. Preparation Build the prefix-sum-on-a-ring helper and write arc(a, b) returning min(clockwise, total - clockwise); test it on a hand-drawn 4-hub ring. Add explicit unit tests for the wrap-around leg (last requested hub back through Hub 1) and the m == 2 ring before trusting the solution.