Design a meeting scheduler

An office has a fixed set of meeting rooms, each with a unique string id. Build MeetingScheduler, which hands out rooms for meetings:

• MeetingScheduler(room_ids: list[str]): the rooms, in any order (at least one, all ids distinct).
• book(start: int, end: int) -> str | None: reserve a room for the half-open interval [start, end) and return its id. Among all rooms that are free for the whole interval, pick the one whose id is smallest in string order ("room10" < "room2"). If no room is free, book nothing and return None.
• schedule(room_id: str) -> list[tuple[int, int]]: the meetings booked in that room, as (start, end) pairs sorted by start. room_id is always one of the rooms.

Intervals are half-open, so a meeting ending at 20 and one starting at 20 don't clash. Times are integers with 0 <= start < end <= 10**9.

s = MeetingScheduler(["b", "a"])
s.book(10, 20)      # "a"
s.book(15, 25)      # "b"   ("a" is busy from 15 to 20)
s.book(20, 30)      # "a"   (back-to-back with 10-20 is fine)
s.book(12, 18)      # None  (both rooms are busy at 15)
s.schedule("a")     # [(10, 20), (20, 30)]
s.schedule("b")     # [(15, 25)]

A room may collect tens of thousands of meetings. Checking whether one room is free must take O(log n) in that room's number of meetings, not a scan of all of them.