Merging, overlap detection, and sweep-line scheduling problems.
3 questions in Data Structures & Algorithms.
Sorting by start time is what reduces the merge to a single scan carrying one scalar, the end of the open block; sorting by end breaks it, and sorting by end is instead correct for maximum non-overlapping selection. Both run in O(n log n), and equal timestamps decide the answer.
Pool every busy interval, sort by start, and sweep with a cursor marking the end of the busy block so far. Any start beyond the cursor opens a common free slot. Use half-open intervals so back-to-back meetings leave no zero-length gap, and clip to the working day. Sorting costs O(N log N), or O(N log k) with a heap.
The answer is the maximum number of meetings running at once, found by sweeping the start and end times in sorted order and tracking a running count. The starts and ends can be sorted independently of each other, and a meeting ending exactly when another begins must free its room or you will overcount.