Meeting Rooms Problem Minimum Rooms Required in DSA C - Execution Trace

Concept Flow - Meeting Rooms Problem Minimum Rooms Required

Sort intervals by start time

↓

Initialize min-heap for end times

↓

For each interval

↓

If min-heap top <= current start

↓

Pop min-heap top

↓

Push current end time

↓

Heap size = rooms needed

↓

Done

Sort meetings by start time, use a min-heap to track earliest ending meeting, reuse room if possible, else add new room.

Execution Sample

DSA C

intervals = [[0,30],[5,10],[15,20]];
sort intervals by start;
minHeap = empty;
for each interval:
  if minHeap.top <= interval.start:
    pop minHeap;
  push interval.end;
rooms = minHeap.size;

Calculate minimum meeting rooms by sorting intervals and using a min-heap to track ongoing meetings.

Execution Table

Step	Operation	Current Interval	Min-Heap Before	Action	Min-Heap After	Rooms Needed	Visual State
1	Sort intervals	[[0,30],[5,10],[15,20]]	[]	Sort by start time	[[0,30],[5,10],[15,20]]	0	[0,30] [5,10] [15,20]
2	Process interval	[0,30]	[]	Push end=30	[30]	1	[30]
3	Process interval	[5,10]	[30]	5 < 30, push end=10	[10,30]	2	[10,30]
4	Process interval	[15,20]	[10,30]	10 <= 15, pop 10, push 20	[20,30]	2	[20,30]
5	End	No more intervals	[20,30]	Heap size=2	[20,30]	2	Minimum rooms = 2

💡 All intervals processed, min-heap size is the minimum rooms required.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
intervals	[[0,30],[5,10],[15,20]]	[[0,30],[5,10],[15,20]]	[[0,30],[5,10],[15,20]]	[[0,30],[5,10],[15,20]]	[[0,30],[5,10],[15,20]]
minHeap	[]	[30]	[10,30]	[20,30]	[20,30]
rooms	0	1	2	2	2

Key Moments - 3 Insights

Why do we pop from the min-heap when the earliest end time is less than or equal to the current start time?

Because the meeting with the earliest end time has finished before the current meeting starts, so we can reuse that room. This is shown in step 4 where 10 <= 15, so we pop 10 and reuse the room.

Why do we push the current meeting's end time into the min-heap even after popping?

We push the current meeting's end time to track when this meeting will finish. This keeps the min-heap updated with ongoing meetings. Step 4 shows popping 10 and pushing 20.

Why does the size of the min-heap represent the number of rooms needed?

Each element in the min-heap represents a meeting currently occupying a room. The heap size shows how many rooms are occupied simultaneously. At the end (step 5), heap size is 2, so 2 rooms are needed.

Visual Quiz - 3 Questions

Look at the execution table at step 3. What is the min-heap state before processing the interval [5,10]?

A []

B [10]

C [30]

D [20,30]

💡 Hint

Check the 'Min-Heap Before' column at step 3 in the execution table.

At which step does the min-heap size increase to 2?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look at the 'Rooms Needed' column in the execution table to see when it changes to 2.

If the interval [5,10] was changed to [35,40], how would the min-heap size after step 3 change?

A It would remain 1

B It would increase to 2

C It would decrease to 0

D It would become 3

💡 Hint

If the new interval starts after the earliest end time, the room can be reused, so heap size does not increase.

Concept Snapshot

Meeting Rooms Minimum Rooms Required:
- Sort intervals by start time
- Use min-heap to track earliest end time
- For each interval:
  - If earliest end <= current start, pop heap
  - Push current end
- Heap size = minimum rooms needed

Full Transcript

To find the minimum number of meeting rooms needed, first sort the meeting intervals by their start times. Then, use a min-heap to keep track of the end times of meetings currently occupying rooms. For each meeting, if the earliest finishing meeting ends before or when the current meeting starts, reuse that room by popping from the heap. Always push the current meeting's end time into the heap. The size of the heap after processing all intervals is the minimum number of rooms required.

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