Recognized it as Meeting Rooms II pretty fast, which was both good and slightly annoying because it felt like they just renamed the problem and called it a day.
Clarify that this is the classic 'minimum meeting rooms' problem, which can be solved by sorting start and end times separately and using a two-pointer sweep to count concurrent intervals. Alternatively, use a min-heap to track end times of ongoing rides, adding a new taxi only when the earliest ending ride is still in progress. Explain the time and space complexity and discuss edge cases like empty input or zero-length intervals.
Pro tip: Mention that the two-pointer approach is often preferred in interviews for its simplicity and O(n log n) time with O(n) space, but be prepared to discuss the heap approach as it naturally handles streaming data. Also, explicitly state that intervals are half-open [start, end) so a ride ending at time t does not conflict with one starting at t.
Confirm that intervals are half-open, ask about input size, and whether the list is sorted. Discuss edge cases such as empty input or zero-length intervals.
Decide between the two-pointer sweep (sort starts and ends separately) or the min-heap method (sort by start time and track end times). Explain why both yield the same result.
For two-pointer: sort starts and ends, iterate through starts, increment count when a start is before the current end, else move the end pointer. For heap: sort by start, push end times into a min-heap, and pop if the earliest end <= current start.
State that both approaches run in O(n log n) time due to sorting and use O(n) space. Test with examples like [[0,30],[5,10],[15,20]] and discuss how the algorithm handles them.
Compare the two-pointer and heap approaches in terms of code simplicity and adaptability to streaming input. Mention that the heap approach can be extended to find the actual assignment of taxis if needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.