Minimum Number of Rooms to Host Overlapping Meetings

You are given a list of meetings, each with a start time and an end time. Every meeting must be held in a room, and a room can host only one meeting at a time. Return the smallest number of rooms that lets every meeting take place at its scheduled time. A meeting `[start, end]` occupies its room over the half-open interval from `start` up to, but not including, `end`. A meeting that ends at time `t` and another that starts at time `t` can therefore use the same room. ### Function Signature ```python def min_rooms(meetings: list[list[int]]) -> int: ``` ### Rules - `meetings[i] = [start_i, end_i]` describes meeting `i`. - Two meetings conflict exactly when their half-open intervals overlap, that is, when `start_a < end_b` and `start_b < end_a`. - Meetings may be given in any order. Identical meetings may appear more than once, and each copy needs its own room. - An empty list needs `0` rooms. - The answer is a single integer, so it is unique. ### Constraints - `0 <= len(meetings) <= 10^5` - `0 <= start_i < end_i <= 10^9` - All values fit in a 32-bit signed integer. ### Examples **Example 1** Input: `meetings = [[9, 12], [10, 11], [11, 13], [10, 14]]` Output: `3` Explanation: during `[10, 11)` three meetings run at once: `[9, 12]`, `[10, 11]` and `[10, 14]`. When `[10, 11]` ends at time 11, `[11, 13]` can take over its room, so no moment ever needs a fourth room. **Example 2** Input: `meetings = [[4, 7], [7, 9], [1, 4]]` Output: `1` Explanation: the meetings run back to back, and each starts exactly when the previous one ends, so a single room is enough. **Example 3** Input: `meetings = []` Output: `0`