Meeting Rooms III

Meeting Rooms III

What is this problem about?

The "Meeting Rooms III" interview question is an advanced scheduling problem that typically asks you to determine which meeting room was used the most, or some other metric related to room usage, given a set of meetings and a fixed number of rooms. Unlike its predecessors, this problem introduces complexities like room availability, duration, and potentially priorities or waiting times. You'll need to allocate meetings to available rooms, and if no room is available, you might need to determine when the next room becomes free, or which room finishes earliest. This problem deeply tests your ability to manage resources and simulate complex events over time.

Why is this asked in interviews?

This "Meeting Rooms III" coding problem is a challenging and frequently asked question in technical interviews at leading tech companies like Uber and Google, designed to evaluate a candidate's mastery of complex data structures and event-driven simulation. It goes far beyond basic interval overlaps, requiring sophisticated management of available resources and efficient querying for the next available slot. Interviewers use this to assess advanced algorithmic thinking, the ability to combine multiple data structures (like heaps and hash tables), and meticulous attention to detail in implementing a simulation. Success demonstrates top-tier problem-solving skills and robust system design intuition.

Algorithmic pattern used

The most effective algorithmic pattern for the "Meeting Rooms III" interview question involves a Simulation approach, heavily relying on Min-Heaps (Priority Queues) for efficient event management.

1. Sort meetings: First, sort all incoming meetings by their start times.
2. Manage room availability: Use two min-heaps:
   • available_rooms: Stores the indices of rooms that are currently free. Initially, this heap contains all room indices.
   • occupied_rooms: Stores pairs (finish_time, room_index) for rooms that are currently occupied, ordered by finish_time.
3. Iterate through sorted meetings: For each meeting:
   • Free up rooms: While occupied_rooms is not empty and its top element's finish_time is less than or equal to the current meeting's start_time, move these rooms from occupied_rooms to available_rooms.
   • Allocate room:
     • If available_rooms is not empty, allocate the room with the smallest index from available_rooms. Calculate its end_time = meeting.start_time + meeting.duration. Add (end_time, room_index) to occupied_rooms.
     • If available_rooms is empty, all rooms are occupied. Find the room in occupied_rooms that finishes earliest (earliest_finish_time, room_index). The current meeting must wait until this room is free. Its start_time becomes earliest_finish_time. Calculate its end_time = earliest_finish_time + meeting.duration. Update the entry in occupied_rooms or move it.
4. Track room usage in a separate array, incrementing counts for the allocated room.

Example explanation

Suppose we have n = 2 rooms and meetings [[0, 10], [1, 5], [2, 12]]. We want to count how many times each room was used.

1. Sort meetings: [[0, 10], [1, 5], [2, 12]]

2. Initialize: available_rooms = [0, 1] (min-heap) occupied_rooms = [] (min-heap of (finish_time, room_index)) room_usage = [0, 0]

3. Process [0, 10]:

   • available_rooms has 0. Take room 0.
   • Add (0 + 10, 0) -> (10, 0) to occupied_rooms.
   • room_usage = [1, 0].

4. Process [1, 5]:

   • occupied_rooms.peek() is (10, 0). Current start_time is 1. 1 < 10, so no room is free.
   • available_rooms has 1. Take room 1.
   • Add (1 + 5, 1) -> (6, 1) to occupied_rooms.
   • room_usage = [1, 1].

5. Process [2, 12]:

   • occupied_rooms.peek() is (6, 1). Current start_time is 2. 2 < 6. No room free yet.
   • available_rooms is empty. Must wait for a room.
   • Smallest finish_time in occupied_rooms is 6 (from room 1).
   • Meeting [2, 12] starts at 6. Its end time becomes 6 + (12-2) = 16.
   • Remove (6, 1) from occupied_rooms. Add (16, 1) to occupied_rooms.
   • room_usage = [1, 2].

Final room usage based on specific problem (e.g., room 0 used 1 time, room 1 used 2 times).

Common mistakes candidates make

A major pitfall in the "Meeting Rooms III" interview question is mishandling the scenario where all rooms are occupied, requiring a meeting to wait. Incorrectly updating the start time of the waiting meeting or failing to re-add the room to the correct heap with its new end time are frequent errors. Another mistake is using inefficient data structures for managing available or occupied rooms, leading to a time complexity worse than O(N log N). Candidates might also struggle with tie-breaking rules, for instance, which room to pick if multiple are available at the same time (e.g., smallest index).

Interview preparation tip

To ace the Meeting Rooms III coding problem, first master its predecessors ("Meeting Rooms" and "Meeting Rooms II"). Then, focus heavily on Simulation using Min-Heaps. You'll need two heaps: one for available rooms (perhaps storing just indices) and another for occupied rooms (storing (finish_time, room_index)). Practice the logic for freeing up rooms, allocating rooms, and, most critically, handling meetings that must wait. Work through various scenarios, manually tracing how meetings are assigned and how the heaps change. Pay attention to tie-breaking rules for room selection. This problem is a comprehensive test of your ability to manage complex events and state.