Meeting Rooms Problem Minimum Rooms Required in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the intervals by their start time.

DSA C#

Array.Sort(intervals, (a, b) => a.Start [1] b.Start);

Drag options to blanks, or click blank then click option'

A-

B>

C<

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' results in incorrect (descending) sorting order.

Using '>' causes incorrect sorting.

Using '==' causes no sorting.

We use a.Start - b.Start which returns negative if a.Start is less than b.Start to sort ascending by start time.

2fill in blank

medium

Complete the code to check if the current meeting starts after the earliest ending meeting.

DSA C#

if (intervals[i].Start [1] minHeap.Peek())

Drag options to blanks, or click blank then click option'

A<=

B<

C>=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '<=' causes wrong logic for room reuse.

Using '>' excludes meetings starting exactly at end time.

If the current meeting starts at or after the earliest ending meeting, we can reuse that room.

3fill in blank

hard

Fix the error in removing the earliest ending meeting from the min heap.

DSA C#

minHeap.[1]();

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ADequeue

BPoll

CRemove

DPop

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Pop', 'Poll', or 'Remove' causes error as they are not methods in C# PriorityQueue.

In C#, PriorityQueue uses 'Dequeue' to remove the smallest element.

4fill in blank

hard

Fill both blanks to add the current meeting's end time to the min heap and update the minimum rooms required.

DSA C#

minHeap.[1](intervals[i].End, intervals[i].End);
minRooms = Math.[2](minRooms, minHeap.Count);

Drag options to blanks, or click blank then click option'

AEnqueue

BMax

CMin

DAdd

Attempts:

3 left

💡 Hint

Common Mistakes

Using Min instead of Max causes incorrect room count.

Using Add instead of Enqueue causes error.

We add the current meeting's end time to the min heap using Enqueue. Then update minRooms with the maximum of current minRooms and minHeap count.

5fill in blank

hard

Fill all three blanks to complete the method that returns the minimum number of meeting rooms required.

DSA C#

public int MinMeetingRooms(Interval[] intervals) {
    if (intervals == null || intervals.Length == 0) return 0;
    Array.Sort(intervals, (a, b) => a.Start [1] b.Start);
    var minHeap = new PriorityQueue<int, int>();
    minHeap.[2](intervals[0].End, intervals[0].End);
    int minRooms = 1;
    for (int i = 1; i < intervals.Length; i++) {
        if (intervals[i].Start [3] minHeap.Peek()) {
            minHeap.Dequeue();
        }
        minHeap.Enqueue(intervals[i].End, intervals[i].End);
        minRooms = Math.Max(minRooms, minHeap.Count);
    }
    return minRooms;
}

Drag options to blanks, or click blank then click option'

A-

B>=

CEnqueue

DDequeue

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '>' in sorting causes wrong order.

Using 'Dequeue' instead of 'Enqueue' to add element causes error.

Using '<=' instead of '>=' in condition causes wrong room reuse logic.

Sort intervals by start time using '-'. Add first meeting end time with Enqueue. Check if current meeting starts at or after earliest end time with '>='.

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