What is the time complexity of Meeting Rooms III?

The optimal solution has a time complexity of O(m log n).

What data structures are used to solve Meeting Rooms III?

This problem uses Array, Hash Table, Sorting, Heap (Priority Queue), Simulation. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Meeting Rooms III?

Insight 1: Sorting the meetings by start time allows processing them in chronological order, which is crucial for simulating the allocation process correctly. Insight 2: Using a min-heap (priority queue) for both available rooms and occupied rooms efficiently manages the allocation and deallocation of rooms based on their availability and end times.

Which companies ask LeetCode 2402. Meeting Rooms III in interviews?

This problem is frequently asked at PayPal, Pinterest.

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Meeting Rooms III - LeetCode 2402 Solution

Meeting Rooms III - Complete Solution Guide

Meeting Rooms III is LeetCode problem 2402, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an integer n . There are n rooms numbered from 0 to n - 1 . You are given a 2D integer array meetings where meetings[i] = [start i , end i ] means that a meeting will be held during the half-closed time interval [start i , end i ) . All the values of start i are unique . Meetings are allocated to rooms in the following manner: Each meeting will take place in the unused room with the lowest number. If there are no available rooms, the meeting will be delayed until a room becomes fre

Detailed Explanation

The problem simulates a meeting room allocation scenario. Given 'n' rooms and a list of meetings with start and end times, the goal is to determine which room hosted the most meetings. Meetings are assigned to the lowest-numbered available room. If no rooms are available, the meeting is delayed until a room becomes free, maintaining the original meeting duration. The room that held the most meetings (and the lowest-numbered one in case of a tie) should be returned.

Solution Approach

The solution uses a simulation approach, processing meetings one by one in chronological order. It employs two priority queues: 'available_rooms' to store available rooms (sorted by room number) and 'occupied_rooms' to store rooms currently in use (sorted by end time). For each meeting, the algorithm first frees up any rooms that have finished their meetings. Then, if available rooms exist, the meeting is assigned to the lowest-numbered available room. Otherwise, the meeting is delayed until the earliest-finishing room becomes free, and the end time is adjusted accordingly. Finally, the room that hosted the most meetings is determined.

Step-by-Step Algorithm

Step 1: Sort the meetings array based on their start times in ascending order.
Step 2: Initialize a min-heap 'available_rooms' with room numbers from 0 to n-1.
Step 3: Initialize a min-heap 'occupied_rooms' to store rooms that are currently occupied, sorted by their end times. The heap will store pairs of (end_time, room_number).
Step 4: Initialize an array 'room_counts' to keep track of the number of meetings held in each room.
Step 5: Iterate through the sorted meetings array.
Step 6: For each meeting, check if any rooms in 'occupied_rooms' have become available (end time <= current meeting's start time). If so, move those rooms to 'available_rooms'.
Step 7: If 'available_rooms' is not empty, assign the meeting to the room with the lowest number from 'available_rooms', add the room and the end time to 'occupied_rooms', and increment the corresponding count in 'room_counts'.
Step 8: If 'available_rooms' is empty, delay the meeting until the earliest-finishing room becomes available. Pop the room from 'occupied_rooms', calculate the new end time for the delayed meeting, push the room and the new end time back into 'occupied_rooms', and increment the corresponding count in 'room_counts'.
Step 9: After processing all meetings, iterate through 'room_counts' to find the room with the most meetings. Return the index of that room.

Key Insights

Insight 1: Sorting the meetings by start time allows processing them in chronological order, which is crucial for simulating the allocation process correctly.
Insight 2: Using a min-heap (priority queue) for both available rooms and occupied rooms efficiently manages the allocation and deallocation of rooms based on their availability and end times.
Insight 3: Handling the case where no rooms are available requires delaying the meeting and updating the end time of the room that becomes available earliest, using the original meeting duration.

Complexity Analysis

Time Complexity: O(m log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Sorting, Heap (Priority Queue), Simulation.

Companies

Asked at: PayPal, Pinterest.

Meeting Rooms III | LeetCode 2402

Hard

Array Hash Table Sorting Heap (Priority Queue)Simulation

Time: O(m log n)Space: O(n)

Solution Code

Available in:

import java.util.Arrays;
import java.util.PriorityQueue;

class Solution {
    public int mostBooked(int n, int[][] meetings) {
        Arrays.sort(meetings, (a, b) -> a[0] - b[0]);

        PriorityQueue<Integer> available_rooms = new PriorityQueue<>();
        for (int i = 0; i < n; i++) {
            available_rooms.offer(i);
        }

        PriorityQueue<long[]> occupied_rooms = new PriorityQueue<>((a, b) -> {
            if (a[0] != b[0]) {
                return Long.compare(a[0], b[0]);
            } else {
                return Long.compare(a[1], b[1]);
            }
        });

        int[] room_counts = new int[n];

        for (int[] meeting : meetings) {
            int start = meeting[0];
            int end = meeting[1];

            while (!occupied_rooms.isEmpty() && occupied_rooms.peek()[0] <= start) {
                long[] room_info = occupied_rooms.poll();
                available_rooms.offer((int) room_info[1]);
            }

            if (!available_rooms.isEmpty()) {
                int room_to_use = available_rooms.poll();
                occupied_rooms.offer(new long[]{(long) end, (long) room_to_use});
                room_counts[room_to_use]++;
            } else {
                long[] room_info = occupied_rooms.poll();
                long earliest_free_time = room_info[0];
                int room_to_use = (int) room_info[1];
                long duration = end - start;
                long new_end_time = earliest_free_time + duration;
                occupied_rooms.offer(new long[]{new_end_time, (long) room_to_use});
                room_counts[room_to_use]++;
            }
        }

        int max_meetings = -1;
        int result_room = -1;
        for (int i = 0; i < n; i++) {
            if (room_counts[i] > max_meetings) {
                max_meetings = room_counts[i];
                result_room = i;
            }
        }

        return result_room;
    }
}

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Problem Info

Description

Problem 2402: Meeting Rooms III - Detailed explanation and solution approaches.

Topics

Array Hash Table Sorting Heap (Priority Queue)Simulation

Companies

PayPalPinterest

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2402

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Sort the meetings array based on their start times in ascending order.
2
Step 2: Initialize a min-heap 'available_rooms' with room numbers from 0 to n-1.
3
Step 3: Initialize a min-heap 'occupied_rooms' to store rooms that are currently occupied, sorted by their end times. The heap will store pairs of (end_time, room_number).
4
Step 4: Initialize an array 'room_counts' to keep track of the number of meetings held in each room.
5
Step 5: Iterate through the sorted meetings array.
6
Step 6: For each meeting, check if any rooms in 'occupied_rooms' have become available (end time <= current meeting's start time). If so, move those rooms to 'available_rooms'.
7
Step 7: If 'available_rooms' is not empty, assign the meeting to the room with the lowest number from 'available_rooms', add the room and the end time to 'occupied_rooms', and increment the corresponding count in 'room_counts'.
8
Step 8: If 'available_rooms' is empty, delay the meeting until the earliest-finishing room becomes available. Pop the room from 'occupied_rooms', calculate the new end time for the delayed meeting, push the room and the new end time back into 'occupied_rooms', and increment the corresponding count in 'room_counts'.
9
Step 9: After processing all meetings, iterate through 'room_counts' to find the room with the most meetings. Return the index of that room.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m log n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Sorting meetings contributes O(m log m). Each meeting involves heap operations on available and occupied rooms (O(log n) each), performed m times. The room_counts array stores n counts (O(n)). The space used by heap structures available_rooms, occupied_rooms are also at most O(n). Since the heap operations are intertwined within iterating meetings, the time complexity of O(m log n) dominates and drives the final time complexity. The storage of rooms and room counts drives the space complexity, thus leading to O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Simulation with Array

Instead of using priority queues, you could use an array to represent available rooms and iterate through the array to find the lowest-numbered available room. For occupied rooms, you could use an array and iterate to find the room that will be available earliest. However, searching the arrays for an available room or the earliest finishing room will take O(n) in the worst case.

Time: O(m*n), where m is the number of meetings and n is the number of rooms.Space: O(n)

Trade-offs: This approach is simpler to implement but has a higher time complexity (O(m*n)) compared to the heap-based approach (O(m log n)). It is suitable for small values of 'n' but becomes inefficient for larger 'n'.

✨

Key Insights

1Insight 1: Sorting the meetings by start time allows processing them in chronological order, which is crucial for simulating the allocation process correctly.
2Insight 2: Using a min-heap (priority queue) for both available rooms and occupied rooms efficiently manages the allocation and deallocation of rooms based on their availability and end times.
3Insight 3: Handling the case where no rooms are available requires delaying the meeting and updating the end time of the room that becomes available earliest, using the original meeting duration.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: When all meetings can be scheduled in available rooms without any delays. The solution correctly assigns meetings to rooms and tracks the room counts.
Edge case 2: When some meetings need to be delayed because all rooms are occupied. The solution correctly delays these meetings and updates the end times accordingly.
Edge case 3: When the number of meetings is much larger than the number of rooms. The solution continues to work correctly by delaying meetings as necessary.
Edge case 4: When there is only one room (n=1). The solution handles this by assigning all meetings (potentially delayed) to that one room and correctly reports that room as the most used.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the meetings by start time. This can lead to incorrect room assignments and delays.

✗Common mistake 2: Not properly managing the priority queues, such as failing to remove rooms from 'occupied_rooms' when they become available or not pushing rooms back into 'available_rooms' after they are freed.

✗Common mistake 3: Using integer data types for end times, which can cause overflow issues if the end times are large. Using long data type is crucial.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation in cloud computing environments, where virtual machines (rooms) are allocated to user requests (meetings). The goal is to maximize resource utilization while minimizing wait times.

🏦Real-world scenario 2: Scheduling appointments in a doctor's office or booking conference rooms in an office building. The algorithm can help efficiently allocate time slots (rooms) to appointments (meetings) while considering availability and priority.

🎮Real-world scenario 3: Managing tasks on a CPU. Different processes(meetings) require different time slots. The operating system uses scheduling algorithms similar to this problem to assign the CPU(room) time to different processes.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 2, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Meeting Rooms III"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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