Given a list of meeting time intervals, return the minimum number of rooms needed to hold them all. A room frees up the minute its meeting ends, so a meeting starting exactly then can reuse it.
Examples
Example 1:
Input: intervals = [[0,30],[5,10],[15,20]]
Output: 2
Example 2:
Input: intervals = [[7,10],[2,4]]
Output: 1
Example 3:
Input: intervals = [[1,5],[5,8],[8,12]]
Output: 1
Explanation: one room, reused twice at the boundary.
Example 4:
Input: intervals = [[1,10],[2,7],[3,19],[8,12],[10,20],[11,30]]
Output: 4
Constraints
1 <= intervals.length <= 10^4
intervals[i] is [start, end] with 0 <= start < end <= 10^6
Prerequisites
Sorting as a setup move — most interval problems are one sort away from a single sweep, and choosing whether to sort by START or by END is the decision that matters.
How to think about it
1. Peak Concurrency by Sweep Optimal
Intuition
Rooms needed equals the largest number of meetings running at the same moment. Detach starts from ends, sort each, and sweep: a start before the next end takes a room, an end at or before the next start frees one. Ends win ties, which is exactly the rule that lets a room be reused the minute it frees.
Algorithm
1. Collect and sort all start times and all end times separately. 2. Walk the starts; before each, release every room whose end is at or before it. 3. Take a room, and track the peak.
Time & Space
Time O(n log n). Space O(n). A min-heap of end times expresses the same sweep.