← Bloomberg Interview Insights
Classic heap problem but I fumbled explaining why a min-heap works here before just coding it.
Start by clarifying the problem: intervals are half-open, and we need the maximum number of concurrent meetings. Then present two solutions: a min-heap approach that tracks end times, and a sweep line approach using sorted start and end times. Compare their time and space complexities, and discuss edge cases.
Pro tip: Mention that the problem is equivalent to finding the maximum overlap at any point, and that the sweep line method can be implemented in O(n log n) time with O(n) space, but the heap approach may be more intuitive. Also, note that if intervals are given as [start, end] with start < end, we can treat them as half-open to avoid counting meetings that end exactly when another starts as overlapping.
Confirm that intervals are half-open (e.g., [start, end)) and that meetings ending at the same time as another starts do not overlap. Ask if the input is sorted or if we can modify it.
Recognize that the minimum number of rooms equals the maximum number of overlapping meetings at any time. This reduces the problem to finding the peak concurrency.
Describe either the min-heap approach: sort by start time, use a heap of end times, and for each meeting, if the earliest ending meeting is free, reuse the room; otherwise, allocate a new room. Or the sweep line approach: separate starts and ends, sort both, and use two pointers to count concurrent meetings.
State that both approaches run in O(n log n) time due to sorting, and O(n) space. Mention that the heap approach uses O(n) space for the heap, while the sweep line uses O(n) for the sorted arrays.
Consider empty input, single meeting, all meetings overlapping, and meetings that are back-to-back. Mention that if the input is already sorted, we can skip sorting and achieve O(n) time with the heap approach.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.