Meeting Rooms II

Given a list of meeting time intervals, return the minimum number of conference rooms required. The canonical sweep / heap problem; Bloomberg likes it in phone screens because the cleanest solution involves either a min-heap of end times or a paired array sweep. Requirements Given an array of meeting time intervals intervals[i] = [start_i, end_i], return the minimum number of conference rooms required to host all meetings. Function signature: int minMeetingRooms(int[][] intervals) Follow-ups: Why does the heap-of-end-times approach work? Sort by start, then for each meeting either reuse the room whose end is earliest (if it has ended) or allocate a new one. Walk through the equivalent sweep-line approach: split into start / end events, sort, scan, track the running count. What if you also need to return which meeting goes into which room? Examples intervals = [[0,30],[5,10],[15,20]] minMeetingRooms -> 2 intervals = [[7,10],[2,4]] minMeetingRooms -> 1 Notes Heap solution: sort intervals by start. Maintain a min-heap of end times. For each interval, if heap.top() <= start, pop (reuse a freed room); always push the new end. The heap size at any point is the room count; return its max. Time O(n log n), space O(n). Sweep-line solution: build two sorted arrays of starts and ends. Walk both with two pointers; on each start, increment the active count if it predates the next end, else advance the end pointer. Track the max. Same O(n log n), space O(n). Both solutions are accepted. The sweep version is slightly more cache-friendly and is preferred when the candidate volunteers it. The follow-up that wants room assignments needs the heap variant; pair each end with the room id. Preparation Implement both the heap and the sweep solutions; argue the trade-off. Practice handling tie cases: a meeting ending at the same time another starts does not require a new room (<= start not < start). Be ready to extend to "can attend all meetings" (LeetCode 252) — same sort, simpler condition.