OA: Round-Trip Mission Schedule

Legacy CodeSignal SDE2 OA problem. Two locations A and B have fixed departure timetables. Each mission is a round trip A→B→A, departing only at the next available departure on each side; each leg takes 100 time units. Given a mission count, return the earliest completion time. Requirements Input: sorted integer arrays A[] and B[] (departure times from A and B respectively), integer m (mission count). Each mission consists of: wait at A until the next departure ≥ current_time, travel for 100 units to B, wait at B until the next departure ≥ arrival_time, travel 100 units back to A. Output: the earliest arrival time back at A after the last mission completes. Examples A = [0, 99, 300, 500], B = [101, 220, 440, 900], m = 2 Mission 1: depart A at 0 → arrive B at 100 → next B departure ≥ 100 is 101 → arrive A at 201 Mission 2: next A departure ≥ 201 is 300 → arrive B at 400 → next B departure ≥ 400 is 440 → arrive A at 540 Output: 540 Notes Two pointers on A and B: advance each pointer to the first departure ≥ current_time. If no departure is available, the mission cannot complete. For each mission, do two pointer advances (one on each list); total time O(m + |A| + |B|). Binary search on each list also works (bisect_left(A, current_time)); slightly less efficient but easier to debug. This is a CodeSignal-era problem; newer Hack2Hire OAs use the prompts above instead. Mostly retained for reference if your loop still uses CodeSignal. Preparation Practice the two-pointer pattern over sorted timetables — a similar shape recurs in Uber Eats batching / fleet scheduling problems.