Meeting Rooms Problem Minimum Rooms Required in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Minimum meeting rooms needed for given intervals

What is the output of the following code that calculates the minimum number of meeting rooms required?

DSA Python

def minMeetingRooms(intervals):
    starts = sorted([i[0] for i in intervals])
    ends = sorted([i[1] for i in intervals])
    s, e = 0, 0
    rooms = 0
    while s < len(intervals):
        if starts[s] < ends[e]:
            rooms += 1
            s += 1
        else:
            e += 1
            s += 1
    return rooms

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

A 0

B 2

C 1

D 3

💡 Hint

Think about how many meetings overlap at the same time.

The code sorts start and end times separately. It increments rooms when a meeting starts before the earliest ending meeting ends. Here, three meetings overlap at some point, so 3 rooms are needed.

❓ Predict Output

intermediate

Output for non-overlapping meetings

What is the output of this code for non-overlapping meeting intervals?

DSA Python

def minMeetingRooms(intervals):
    starts = sorted([i[0] for i in intervals])
    ends = sorted([i[1] for i in intervals])
    s, e = 0, 0
    rooms = 0
    while s < len(intervals):
        if starts[s] < ends[e]:
            rooms += 1
            s += 1
        else:
            e += 1
            s += 1
    return rooms

intervals = [[1, 2], [3, 4], [5, 6]]
print(minMeetingRooms(intervals))

A 1

B 2

C 3

D 0

💡 Hint

No meetings overlap here.

Since no meetings overlap, only one room is needed at any time.

🔧 Debug

advanced

Identify the error in meeting rooms calculation

What error does this code raise when run?

DSA Python

def minMeetingRooms(intervals):
    starts = sorted([i[0] for i in intervals])
    ends = sorted([i[1] for i in intervals])
    s, e = 0, 0
    rooms = 0
    while s < len(intervals):
        if starts[s] <= ends[e]:
            rooms += 1
            s += 1
        else:
            e += 1
    return rooms

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

A 0

B IndexError

C 3

D TypeError

💡 Hint

Check the loop increments carefully.

The code increments s only in the if branch but not in else, causing an infinite loop or IndexError when s exceeds length.

🧠 Conceptual

advanced

Why sorting start and end times separately helps?

Why do we sort start and end times separately when calculating minimum meeting rooms?

A To compare meetings in chronological order and track overlaps efficiently

B To find the longest meeting duration

C To count total number of meetings only

D To merge intervals into one big interval

💡 Hint

Think about how to detect overlaps by comparing start and end times.

Sorting start and end times separately allows us to iterate through meetings in order and count how many overlap at any time, which determines rooms needed.

❓ Predict Output

expert

Minimum rooms for complex overlapping intervals

What is the output of this code for complex overlapping intervals?

DSA Python

def minMeetingRooms(intervals):
    starts = sorted([i[0] for i in intervals])
    ends = sorted([i[1] for i in intervals])
    s, e = 0, 0
    rooms = 0
    max_rooms = 0
    while s < len(intervals):
        if starts[s] < ends[e]:
            rooms += 1
            max_rooms = max(max_rooms, rooms)
            s += 1
        else:
            rooms -= 1
            e += 1
    return max_rooms

intervals = [[1, 10], [2, 7], [3, 19], [8, 12], [10, 20], [11, 30]]
print(minMeetingRooms(intervals))

A 4

B 3

C 5

D 6

💡 Hint

Count maximum overlaps at any time.

The maximum number of overlapping meetings is 5 at some point, so 5 rooms are needed.