← Bytedance Interview Insights
I knew this problem but fumbled the heap approach mid-explanation and ended up describing the two-pointer sort trick instead.
Clarify the problem constraints, then propose a solution using a min-heap to track meeting end times. Sort intervals by start time, iterate through them, and for each meeting, if the earliest ending meeting has ended, reuse that room; otherwise allocate a new room. The heap size at the end gives the minimum number of rooms.
Pro tip: Mention that this is equivalent to finding the maximum number of overlapping intervals, and that the heap approach is optimal with O(n log n) time. Also, discuss edge cases like empty input and back-to-back meetings (where end time equals start time) to show thoroughness.
Restate the problem to ensure clarity: given a list of intervals, determine the minimum number of rooms required so that no two meetings overlap. Confirm whether intervals are half-open (e.g., [start, end)) and if back-to-back meetings are allowed.
Decide between sorting-based approaches (e.g., sweep line) or heap-based. Explain that sorting by start time and using a min-heap of end times efficiently tracks room availability.
Describe step-by-step: sort intervals by start time; initialize a min-heap; for each interval, if the heap is not empty and the earliest end time <= current start, pop the heap; push the current end time; the heap size is the answer.
State that sorting takes O(n log n) and each heap operation takes O(log n), leading to O(n log n) overall time and O(n) space in the worst case.
Run through a simple example, such as [[0,30],[5,10],[15,20]], to demonstrate the algorithm and verify the output (2 rooms). Also consider edge cases like empty input or all meetings overlapping.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.