My first instinct was greedy, assign each truck to whichever bay finishes earliest.
This is a scheduling problem where we need to assign trucks to bays in order, minimizing the number of bays such that the total time (makespan) does not exceed the limit. The optimal strategy is to use a greedy algorithm with a min-heap to track the earliest available time of each bay: for each truck, if the earliest available bay can start immediately (i.e., its available time + truck time ≤ limit), assign it there; otherwise, open a new bay. This minimizes the number of bays because we always reuse the bay that becomes free earliest, which is optimal for minimizing the maximum completion time.
Pro tip: Clarify that the greedy approach is optimal by relating it to the classic 'minimum number of machines to schedule jobs by a deadline' problem, and mention that a binary search on the number of bays could also work but is less efficient. Also, handle edge cases like a single truck exceeding the limit by returning -1 or indicating impossibility.
Restate the problem: trucks must be unloaded in order, each bay processes one truck at a time, and we need the minimum number of bays so that every truck finishes within the given total turnaround time. Clarify that the total turnaround time is the maximum completion time across all trucks.
Use a min-heap to track the earliest available time of each bay. For each truck, if the earliest available bay can start immediately and finish within the limit, assign it there and update the bay's available time; otherwise, open a new bay. This greedy strategy minimizes the number of bays.
Demonstrate with a small example, such as unload times [10, 20, 30] and limit 40. Show how the heap evolves and how bays are assigned, ensuring the candidate can explain the process clearly.
Explain that each truck is processed once, and heap operations take O(log k) time where k is the number of bays, leading to O(n log k) time and O(k) space. Mention that k ≤ n, so worst-case O(n log n).
Address cases like a single truck exceeding the limit (return -1), all trucks fitting in one bay, or many trucks requiring many bays. Optionally, mention that binary search on the number of bays with a feasibility check is another approach but may be less efficient.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.