Meeting Rooms Problem Minimum Rooms Required in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Python

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

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

A0

B[0]

C[1]

D[

Attempts:

3 left

💡 Hint

Common Mistakes

Using x[1] instead of x[0] to sort by end time.

Forgetting to use the key parameter in sort.

The intervals are sorted by the first element (start time) using x[0].

2fill in blank

medium

Complete the code to initialize a min-heap for tracking end times.

DSA Python

import heapq
heap = []
heapq.[1](heap, intervals[0][1])

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

Aheappush

Bpush

Cinsert

Dadd

Attempts:

3 left

💡 Hint

Common Mistakes

Using push instead of heappush causes AttributeError.

Using append instead of heappush breaks heap property.

Use heapq.heappush to add an element to the heap.

3fill in blank

hard

Fix the error in the condition to check if the current meeting can reuse a room.

DSA Python

if intervals[i][0] [1] heap[0]:

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

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes wrong logic and incorrect room count.

Using '<' misses the case when meetings start exactly at end time.

The current meeting start time must be greater or equal to the earliest end time to reuse the room.

4fill in blank

hard

Fill both blanks to pop the earliest end time and push the current meeting's end time.

DSA Python

heapq.[1](heap)
heapq.[2](heap, intervals[i][1])

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

Aheappop

Bheappush

Cpop

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop or push instead of heappop/heappush breaks heap property.

Forgetting to pop before pushing causes incorrect room count.

Use heappop to remove the earliest end time and heappush to add the current meeting's end time.

5fill in blank

hard

Fill all three blanks to complete the function that returns minimum meeting rooms required.

DSA Python

def minMeetingRooms(intervals):
    if not intervals:
        return 0
    intervals.sort(key=lambda x: x[[1]])
    import heapq
    heap = []
    heapq.heappush(heap, intervals[0][1])
    for i in range(1, len(intervals)):
        if intervals[i][0] [2] heap[0]:
            heapq.heappop(heap)
        heapq.heappush(heap, intervals[i][[3]])
    return len(heap)

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

A[

B>=

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong indices for start or end times.

Using '<=' instead of '>=' in condition.

Pushing start time instead of end time into heap.

Sort by start time using x[0], check if start >= earliest end time, and push the end time at index 1.

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