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.

intermediate

Which data structure is commonly used to keep track of ongoing meetings in the Meeting Rooms Problem?

A min-heap (priority queue) is used to track the earliest finishing meeting to efficiently allocate rooms.

beginner

Why do we sort the meetings by their start time in the Meeting Rooms Problem?

Sorting by start time helps process meetings in chronological order to allocate rooms as meetings begin.

intermediate

What does it mean if the start time of the current meeting is greater than or equal to the earliest finishing meeting in the heap?

It means a room has freed up, so we can reuse that room for the current meeting by updating the finishing time in the heap.

intermediate

What is the time complexity of the optimal solution for the Meeting Rooms Problem Minimum Rooms Required?

O(n log n), where n is the number of meetings, due to sorting and heap operations.

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

A Count overlapping meetings directly

B Sort meetings by end time

C Use a stack to track meetings

D Sort meetings by start time

Sorting meetings by start time allows processing them in chronological order.

Which data structure helps efficiently find the earliest finishing meeting?

A Queue

B Stack

C Min-heap

D Hash map

A min-heap keeps track of the earliest finishing meeting at the top.

If a meeting starts after the earliest finishing meeting ends, what should we do?

A Add a new room

B Reuse the freed room by updating the heap

C Ignore the meeting

D Remove all meetings from the heap

We can reuse the room by updating the finishing time in the heap.

What does the size of the min-heap represent during the algorithm?

A Number of rooms currently in use

B Number of meetings processed

C Total number of meetings

D Number of meetings finished

The heap size shows how many rooms are occupied at that time.

What is the minimum number of rooms needed if no meetings overlap?

A One

B Two

C Number of meetings

D Zero

If no meetings overlap, one room is enough to hold all meetings sequentially.

Explain step-by-step how to find the minimum number of meeting rooms required given a list of meeting intervals.

Think about processing meetings in order and tracking when rooms free up.

Describe why a min-heap is useful in the Meeting Rooms Problem and how it helps optimize room allocation.

Focus on how to quickly find the earliest finishing meeting.

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

B. The maximum duration of a single meeting

C. The total number of meetings scheduled

D. The earliest start time among all meetings

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 asked

   We want to find how many rooms are needed to schedule all meetings without time conflicts.

3. Final Answer:

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

4. Quick Check:

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

Hint: Count max overlapping meetings to find minimum rooms [OK]

Common Mistakes:

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

2. Which of the following C code snippets correctly declares an array of meeting intervals where each interval has a start and end time?

easy

A. int intervals[2][3] = {{0, 5, 15}, {30, 10, 20}};

B. int intervals[3] = {0, 30, 5, 10, 15, 20};

C. int intervals[3][2] = {0, 30, 5, 10, 15, 20};

D. int intervals[3][2] = {{0, 30}, {5, 10}, {15, 20}};

Solution

1. Step 1: Check array dimensions and initialization

   We need a 2D array with 3 intervals, each having 2 values (start and end).

2. Step 2: Validate each option

   int intervals[3][2] = {{0, 30}, {5, 10}, {15, 20}}; correctly declares int intervals[3][2] with proper nested braces for each interval.

3. Final Answer:

   int intervals[3][2] = {{0, 30}, {5, 10}, {15, 20}}; -> Option D

4. Quick Check:

   Correct 2D array syntax for intervals [OK]

Hint: Use int arr[n][2] for intervals with start and end [OK]

Common Mistakes:

• Using 1D array for 2D interval data
• Incorrect brace nesting in initialization
• Swapping rows and columns in declaration

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

int intervals[3][2] = {{0, 30}, {5, 10}, {15, 20}};
// Assume a function minMeetingRooms(intervals, 3) returns minimum rooms needed
int rooms = minMeetingRooms(intervals, 3);
printf("%d\n", rooms);

Assuming minMeetingRooms correctly implements the meeting rooms logic, what is printed?

medium

A. 1

B. 3

C. 2

D. 0

Solution

1. Step 1: Analyze meeting intervals for overlaps

   Intervals: [0,30], [5,10], [15,20]. The first and second overlap (5 < 30), so need at least 2 rooms.

2. Step 2: Check if any three overlap simultaneously

   Third interval [15,20] overlaps with first [0,30], but second ends at 10, so max overlap is 2 meetings at once.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Max overlapping meetings = 2 [OK]

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

Common Mistakes:

• Assuming all meetings overlap simultaneously
• Counting total meetings instead of overlaps
• Ignoring partial overlaps

4. The following C function aims to find the minimum number of meeting rooms required but has a bug. What is the bug?

int minMeetingRooms(int intervals[][2], int n) {
    int rooms = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (intervals[i][0] < intervals[j][1] && intervals[j][0] < intervals[i][1]) {
                rooms++;
            }
        }
    }
    return rooms;
}

medium

A. It counts overlapping pairs but does not track maximum overlaps for rooms

B. It uses wrong indices in the loops causing out-of-bounds access

C. It returns zero always because rooms is never incremented

D. It sorts intervals incorrectly before counting overlaps

Solution

1. Step 1: Understand the nested loops and condition

   The code counts each overlapping pair by checking if intervals overlap and increments rooms.

2. Step 2: Identify why counting pairs is incorrect

   Counting pairs does not give the minimum rooms needed; we need the maximum number of simultaneous overlaps.

3. Final Answer:

   It counts overlapping pairs but does not track maximum overlaps for rooms -> Option A

4. Quick Check:

   Counting pairs ≠ max simultaneous overlaps [OK]

Hint: Track max overlaps, not just count pairs [OK]

Common Mistakes:

• Confusing pair counts with room count
• Not sorting intervals before processing
• Ignoring simultaneous overlaps

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 meeting rooms needed?

hard

A. 2

B. 3

C. 4

D. 5

Solution

1. Step 1: Sort intervals by start time

   Sorted intervals: [1,4], [2,5], [3,6], [7,9], [10,11]

2. Step 2: Find maximum overlapping meetings

   Between 1-6, three meetings overlap: [1,4], [2,5], [3,6]. After 6, meetings do not overlap.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Max overlaps = 3 rooms needed [OK]

Hint: Sort intervals, count max overlaps for rooms [OK]

Common Mistakes:

• Not sorting intervals before counting
• Counting total meetings instead of overlaps
• Missing overlapping intervals in middle