Meeting Rooms Problem Minimum Rooms Required in DSA Python - 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

How do you represent meetings in the Meeting Rooms Problem?

Meetings are represented as intervals with a start time and an end time, e.g., [start, end].

intermediate

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

Sorting by start time helps process meetings in chronological order to efficiently check overlaps and allocate rooms.

intermediate

What data structure is commonly used to track ongoing meetings in the Meeting Rooms Problem?

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

advanced

Explain the process of allocating rooms using a min-heap in the Meeting Rooms Problem.

For each meeting, compare its start time with the earliest ending meeting in the min-heap. If the current meeting starts after or exactly when the earliest meeting ends, reuse that room by updating the end time in the heap. Otherwise, add a new room by pushing the meeting's end time. The maximum size of the heap during the process is the minimum rooms required.

What does the minimum number of meeting rooms depend on?

A The maximum number of overlapping meetings

B The total number of meetings

C The earliest meeting start time

D The longest meeting duration

The minimum rooms needed equals the maximum number of meetings overlapping at any time.

Which data structure helps efficiently find the earliest finishing meeting?

A Stack

B Queue

C Min-heap

D Hash map

A min-heap keeps track of the smallest end time at the top, helping to free rooms quickly.

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

A Sort meetings by start time

B Count total meetings

C Add all meetings to a heap

D Sort meetings by end time

Sorting by start time allows processing meetings in chronological order.

If a meeting starts exactly when another ends, can they use the same room?

A Only if the meetings are on different days

B No

C Only if the meetings are short

D Yes

Meetings that start when another ends do not overlap and can share the same room.

What does the maximum size of the min-heap represent in the algorithm?

A Total meetings

B Minimum rooms required

C Maximum meeting duration

D Number of free rooms

The maximum heap size shows the peak number of rooms occupied simultaneously, which is the minimum rooms required.

Describe the step-by-step approach to find the minimum number of meeting rooms required for a list of meetings.

Think about how to track overlapping meetings efficiently.

Explain why sorting meetings by start time is important in the Meeting Rooms Problem.

Consider how order affects checking meeting conflicts.

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. Final Answer:

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

3. Quick Check:

   Minimum rooms = rooms needed to avoid overlap [OK]

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

Common Mistakes:

• Confusing total meetings with rooms needed
• Ignoring overlapping intervals
• Counting meeting duration instead of rooms

2. Which of the following Python code snippets correctly sorts a list of meeting intervals by their start time?

easy

A. intervals.sort(key=lambda x: x[1])

B. intervals.sort(key=lambda x: x[0])

C. intervals.sort()

D. intervals.sort(key=lambda x: x)

Solution

1. Step 1: Identify sorting by start time

   Each interval is a list or tuple like [start, end]. Sorting by x[0] sorts by start time.

2. Step 2: Check other options

   Sorting by x[1] sorts by end time, sort() without key sorts lexicographically (primarily by start but also end), lambda x: x sorts by whole interval.

3. Final Answer:

   intervals.sort(key=lambda x: x[0]) -> Option B

4. Quick Check:

   Sort by start time = key=lambda x: x[0] [OK]

Hint: Sort intervals by start time using key=lambda x: x[0] [OK]

Common Mistakes:

• Sorting by end time instead of start time
• Using sort() without key on lists of lists
• Confusing lambda syntax

3. What is the output of the following Python code that calculates minimum meeting rooms needed?

intervals = [[0, 30], [5, 10], [15, 20]]
starts = sorted([i[0] for i in intervals])
ends = sorted([i[1] for i in intervals])
rooms = 0
end_ptr = 0
for start in starts:
    if start < ends[end_ptr]:
        rooms += 1
    else:
        end_ptr += 1
print(rooms)

medium

A. 1

B. 3

C. 2

D. 0

Solution

1. Step 1: Sort start and end times

   starts = [0,5,15], ends = [10,20,30]. Initialize rooms=0, end_ptr=0.

2. Step 2: Trace the loop

   - start=0 < ends[0]=10: yes, rooms=1 (end_ptr=0)
   - start=5 < ends[0]=10: yes, rooms=2 (end_ptr=0)
   - start=15 < ends[0]=10: no, end_ptr=1
   Final rooms=2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Maximum concurrent meetings = 2 [OK]

Hint: Compare starts with earliest end to count rooms [OK]

Common Mistakes:

• Not moving end pointer correctly
• Counting rooms too many times
• Mixing start and end comparisons

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

def minMeetingRooms(intervals):
    starts = sorted([i[0] for i in intervals])
    ends = sorted([i[1] for i in intervals])
    rooms = 0
    end_ptr = 0
    for start in starts:
        if start > ends[end_ptr]:
            end_ptr += 1
        else:
            rooms += 1
    return rooms

print(minMeetingRooms([[0, 30], [5, 10], [15, 20]]))

medium

A. The comparison should be start < ends[end_ptr], not start > ends[end_ptr]

B. The ends list is not sorted

C. The rooms variable is not initialized

D. The function does not return any value

Solution

1. Step 1: Analyze the comparison logic

   The code uses if start > ends[end_ptr] to advance end_ptr, but correct logic requires advancing when start >= ends[end_ptr] (or equivalently, increment rooms when start < ends[end_ptr]).

2. Step 2: Understand impact

   Wrong comparison fails when start == ends[end_ptr], incorrectly counting extra room.

3. Final Answer:

   The comparison should be start < ends[end_ptr], not start > ends[end_ptr] -> Option A

4. Quick Check:

   Correct comparison fixes room count [OK]

Hint: Use start < ends[end_ptr] to check overlap [OK]

Common Mistakes:

• Using wrong comparison operator
• Not sorting ends list
• Forgetting to increment end pointer

5. Given the meeting intervals [[1, 4], [2, 5], [7, 9], [3, 6], [10, 11]], what is the minimum number of meeting rooms required to schedule all meetings without overlap?

hard

A. 4

B. 2

C. 1

D. 3

Solution

1. Step 1: Sort start and end times

   starts = [1,2,3,7,10], ends = [4,5,6,9,11].

2. Step 2: Simulate two-pointer algorithm

   rooms=0, end_ptr=0
   - start=1 < 4: rooms=1
   - start=2 < 4: rooms=2
   - start=3 < 4: rooms=3
   - start=7 < 4: no, end_ptr=1
   - start=10 < 5: no, end_ptr=2
   The final rooms value (3) is the max concurrent meetings needed.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Max overlap (e.g., [1,4],[2,5],[3,6]) = 3 [OK]

Hint: Track overlaps by comparing starts and ends [OK]

Common Mistakes:

• Not updating end pointer correctly
• Counting all meetings as overlapping
• Ignoring meeting end times