Meeting Rooms Problem Minimum Rooms Required in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Meeting Rooms Problem Minimum Rooms Required?

To find the minimum number of meeting rooms needed to schedule all given meetings without overlap.

beginner

In the Meeting Rooms problem, why do we sort the intervals by start time?

Sorting by start time helps us process meetings in the order they begin, making it easier to track overlaps and allocate rooms efficiently.

intermediate

What data structure is commonly used to track the end times of meetings currently occupying rooms?

A min-heap (priority queue) is used to keep track of the earliest finishing meeting, so we can reuse rooms when possible.

intermediate

Explain the step when a meeting's start time is greater or equal to the earliest end time in the min-heap.

If the current meeting starts after or when the earliest meeting ends, we can reuse that room by removing the earliest end time from the heap and adding the current meeting's end time.

beginner

What does the maximum size of the min-heap represent during processing?

It represents the minimum number of meeting rooms required to hold all meetings without conflicts.

What is the first step in solving the Meeting Rooms Minimum Rooms Required problem?

A Sort meetings by start time

B Sort meetings by end time

C Use a stack to track meetings

D Count total meetings

Sorting meetings by start time allows us to process them in chronological order.

Which data structure helps efficiently find the earliest finishing meeting?

A Min-heap (priority queue)

B Stack

C Queue

D HashSet

A min-heap keeps the smallest end time at the top, helping to track room availability.

If a meeting starts before the earliest meeting ends, what must we do?

A Remove the earliest meeting from the heap

B Reuse the same room

C Allocate a new room

D Ignore the meeting

We need a new room because the current meeting overlaps with an ongoing one.

What does adding a meeting's end time to the min-heap represent?

A Ending all meetings

B Freeing a room

C Starting a new meeting

D Allocating a room for that meeting

Adding the end time means the room is occupied until that time.

The minimum number of rooms needed equals:

A Total meetings

B Maximum size of the min-heap

C Number of meetings that start first

D Number of meetings that end last

The min-heap size shows how many rooms are occupied simultaneously; the maximum size determines the minimum rooms required.

Describe the approach to find the minimum number of meeting rooms required for a list of meeting intervals.

Think about sorting and tracking meeting end times.

Explain why a min-heap is used in the Meeting Rooms Minimum Rooms Required problem and how it helps optimize room allocation.

Focus on how to efficiently find the earliest meeting that frees a room.

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