My first instinct was to sort by start time and just greedily assign an available car if one exists, otherwise spin up a new one.
Model each rental as an interval and recognize that the minimum number of cars equals the maximum number of overlapping intervals at any point in time. Sort all start and end times, then sweep through them to count concurrent rentals, updating the maximum. This yields an O(n log n) solution.
Pro tip: Clarify upfront whether a rental ending at time t and another starting at t can share the same car; if so, process end times before start times when sweeping to avoid overcounting. Also mention that if actual assignments are needed, a min-heap of car availability times can produce them.
Ask whether intervals are half-open (end == start allowed) and whether actual car assignments are needed or just the count. Confirm input format and constraints.
Explain that minimizing cars is equivalent to finding the maximum number of rentals active at any moment, since each concurrent rental needs a distinct car.
Create events for each start (+1) and end (-1), sort them by time with ends before starts at equal times, then sweep to track current active rentals and the maximum.
State O(n log n) time and O(n) space. If assignments are required, describe using a min-heap of car end times to reuse cars efficiently.
Walk through a small example (e.g., intervals [1,4], [2,5], [3,6]) to verify the count and assignment logic, including boundary cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.