The core idea clicked pretty fast for me, sort by start time and use a min-heap to track when each car becomes free.
Model each rental as an interval [pickup, return] and recognize that the minimum number of cars equals the maximum number of overlapping intervals at any point in time. Use a sweep-line algorithm to compute this maximum, then assign rentals to cars via a greedy interval partitioning approach, ensuring no car is double-booked.
Pro tip: Clarify whether a car can be reassigned immediately after a return (i.e., if one rental ends at time t and another starts at t, they can share a car). This edge case often changes the answer and shows attention to detail.
Confirm whether intervals are inclusive/exclusive at endpoints, if times are integers or continuous, and if a car can be reused instantly after return. Discuss handling of empty input or invalid intervals.
Create events for each pickup (+1) and return (-1), sort by time (with returns before pickups if instant reuse allowed), and track the running sum to find the maximum concurrent rentals. This maximum is the minimum number of cars needed.
Sort rentals by pickup time. Use a min-heap of cars keyed by their next available time (return time). For each rental, if the earliest available car is free by the pickup time, assign it and update its availability; otherwise, allocate a new car.
Explain that sorting takes O(n log n) and heap operations take O(n log n), so overall O(n log n) time and O(n) space. Argue correctness: the sweep-line gives a lower bound, and the greedy assignment achieves it without conflicts.
Output the number of cars (from step 2 or the number of cars used in step 3) and a mapping from each car to its list of assigned rentals, ensuring all rentals are covered exactly once.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.