Meeting Rooms Problem Minimum Rooms Required in DSA C# - Build from Scratch

Meeting Rooms Problem Minimum Rooms Required

📖 Scenario: You are organizing a conference with multiple meetings scheduled at different times. You want to find out the minimum number of meeting rooms needed so that no meetings overlap in the same room.

🎯 Goal: Build a program that takes a list of meeting time intervals and calculates the minimum number of meeting rooms required to hold all meetings without conflicts.

📋 What You'll Learn

Create a list of meeting intervals with start and end times

Sort the meetings by their start times

Use a data structure to track ongoing meetings and find overlaps

Calculate and print the minimum number of rooms needed

💡 Why This Matters

🌍 Real World

Scheduling meeting rooms in offices, conference centers, or classrooms to avoid conflicts.

💼 Career

Useful for software engineers working on calendar apps, booking systems, or resource management tools.

1

Create the list of meeting intervals

Create a list called meetings of tuples with these exact intervals: (0, 30), (5, 10), (15, 20).

DSA C#

# Your code here

Hint

Use List<(int, int)> and add the exact tuples inside curly braces.

2

Sort the meetings by start time

Add code to sort the meetings list by the start time of each meeting using meetings.Sort((a, b) => a.start.CompareTo(b.start)).

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        List meetings = new List
        {
            (0, 30),
            (5, 10),
            (15, 20)
        };
        // Sort meetings by start time
        // Your code here
    }
}

Hint

Use the Sort method with a lambda comparing a.start and b.start.

3

Calculate minimum number of meeting rooms

Create a List<int> called endTimes to track meeting end times. Loop through each meeting in meetings. For each meeting, if endTimes is not empty and the earliest end time is less than or equal to the current meeting's start, remove that end time. Then add the current meeting's end time to endTimes. Use endTimes.Sort() after each addition.

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        List meetings = new List
        {
            (0, 30),
            (5, 10),
            (15, 20)
        };
        meetings.Sort((a, b) => a.start.CompareTo(b.start));

        // Calculate minimum rooms
        // Your code here
    }
}

Hint

Use a list to keep track of end times. Remove the earliest end time if it is free before the current meeting starts. Add the current meeting's end time and sort the list.

4

Print the minimum number of meeting rooms required

Print the count of endTimes using Console.WriteLine(endTimes.Count) to show the minimum number of meeting rooms needed.

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        List meetings = new List
        {
            (0, 30),
            (5, 10),
            (15, 20)
        };
        meetings.Sort((a, b) => a.start.CompareTo(b.start));

        List endTimes = new List();

        foreach (var meeting in meetings)
        {
            if (endTimes.Count > 0 && endTimes[0] <= meeting.start)
            {
                endTimes.RemoveAt(0);
            }
            endTimes.Add(meeting.end);
            endTimes.Sort();
        }

        // Print minimum rooms
        // Your code here
    }
}

Hint

The minimum number of rooms is the size of the endTimes list after processing all meetings.

Practice

1. What does the "Meeting Rooms Problem Minimum Rooms Required" ask you to find given a list of meeting time intervals?

easy

A. The maximum duration of a single meeting

B. The earliest start time among all meetings

C. The total number of meetings scheduled

D. The minimum number of meeting rooms needed to hold all meetings without overlap

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum number of rooms needed so that no two meetings overlap in the same room.

2. Step 2: Clarify what is being counted

   We count rooms, not meetings or durations. The goal is to avoid time conflicts in the same room.

3. Final Answer:

   The minimum number of meeting rooms needed to hold all meetings without overlap -> Option D

4. Quick Check:

   Minimum rooms = no overlapping meetings at the same time [OK]

Hint: Count overlaps to find minimum rooms needed [OK]

Common Mistakes:

• Confusing number of meetings with number of rooms
• Thinking about meeting durations instead of overlaps
• Ignoring overlapping intervals

2. Which of the following C# code snippets correctly initializes a list of meeting intervals represented as tuples of start and end times?

easy

A. var meetings = new List<(int, int)> { 9, 10, 11, 12 };

B. var meetings = new List { [9, 10], [11, 12] };

C. var meetings = new List<(int start, int end)> { (9, 10), (11, 12) };

D. var meetings = new List<(int start, int end)> { (start: 9, end: 10), (start: 11) };

Solution

1. Step 1: Check tuple list initialization syntax

   var meetings = new List<(int start, int end)> { (9, 10), (11, 12) }; correctly uses List of tuples with named elements and proper tuple syntax.

2. Step 2: Identify errors in other options

   var meetings = new List { [9, 10], [11, 12] }; uses invalid array syntax inside list initialization. var meetings = new List<(int, int)> { 9, 10, 11, 12 }; tries to add ints directly, not tuples. var meetings = new List<(int start, int end)> { (start: 9, end: 10), (start: 11) }; misses end value in second tuple.

3. Final Answer:

   var meetings = new List<(int start, int end)> { (9, 10), (11, 12) }; -> Option C

4. Quick Check:

   Correct tuple list syntax = var meetings = new List<(int start, int end)> { (9, 10), (11, 12) }; [OK]

Hint: Use (int, int) tuples with curly braces for intervals [OK]

Common Mistakes:

• Using square brackets instead of parentheses for tuples
• Missing tuple elements or names
• Adding ints instead of tuples

3. Given the following C# code, what is the output?

var intervals = new List<(int start, int end)>{ (0, 30), (5, 10), (15, 20) };
int minRooms = MinMeetingRooms(intervals);
Console.WriteLine(minRooms);

int MinMeetingRooms(List<(int start, int end)> intervals) {
    var starts = intervals.Select(i => i.start).OrderBy(t => t).ToList();
    var ends = intervals.Select(i => i.end).OrderBy(t => t).ToList();
    int rooms = 0, endIndex = 0;
    foreach (var start in starts) {
        if (start < ends[endIndex]) rooms++;
        else endIndex++;
    }
    return rooms;
}

medium

A. 3

B. 2

C. 1

D. 0

Solution

1. Step 1: Sort start and end times

   Starts = [0, 5, 15], Ends = [10, 20, 30]

2. Step 2: Iterate starts comparing with ends

   start=0 < 10 -> rooms=1; start=5 < 10 -> rooms=2; start=15 >= 10 -> endIndex=1, no increment rooms; final rooms=2

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Overlapping intervals = 2 rooms needed [OK]

Hint: Sort starts and ends, count overlaps with end pointer [OK]

Common Mistakes:

• Not sorting start and end times separately
• Incrementing rooms incorrectly when meetings don't overlap
• Confusing start and end comparisons

4. The following code attempts to find the minimum number of meeting rooms needed. What is the bug?

int MinRooms(List<(int start, int end)> intervals) {
    intervals.Sort((a, b) => a.start.CompareTo(b.start));
    var heap = new SortedSet<int>();
    foreach (var interval in intervals) {
        if (heap.Count > 0 && interval.start >= heap.Min) {
            heap.Remove(heap.Min);
        }
        heap.Add(interval.end);
    }
    return heap.Count;
}

medium

A. SortedSet does not allow duplicate end times, causing incorrect room count

B. Intervals are not sorted by end time before processing

C. The condition should be interval.start > heap.Min instead of >= to avoid overlap

D. Heap should store start times instead of end times

Solution

1. Step 1: Understand SortedSet behavior

   SortedSet does not allow duplicates, so multiple meetings ending at same time cause missing removals.

2. Step 2: Identify impact on room count

   Without duplicates, heap may keep extra end times, inflating room count incorrectly.

3. Final Answer:

   SortedSet does not allow duplicate end times, causing incorrect room count -> Option A

4. Quick Check:

   Data structure must allow duplicates for correct room count [OK]

Hint: Use a min-heap that supports duplicates, not SortedSet [OK]

Common Mistakes:

• Using SortedSet instead of a priority queue or min-heap
• Confusing start and end time comparisons
• Incorrect condition for removing from heap

5. You have the following meeting intervals: [(1, 4), (2, 5), (7, 9), (3, 6), (10, 11)]. Using the minimum rooms required approach, what is the minimum number of rooms needed to schedule all meetings without overlap?

hard

A. 3

B. 2

C. 4

D. 1

Solution

1. Step 1: Sort start and end times

   Starts = [1, 2, 3, 7, 10], Ends = [4, 5, 6, 9, 11]

2. Step 2: Count overlaps using two pointers

   start=1 < 4 rooms=1; start=2 < 4 rooms=2; start=3 < 4 rooms=3; start=7 >= 4 endIndex=1 no increment rooms; start=10 >= 5 endIndex=2 no increment rooms; final rooms=3

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Maximum overlaps = 3 rooms needed [OK]

Hint: Sort starts and ends, increment rooms on overlap [OK]

Common Mistakes:

• Not sorting intervals before processing
• Counting rooms incorrectly when meetings end
• Ignoring overlapping intervals