Meeting Rooms Problem Minimum Rooms Required in DSA Python - Time & Space Complexity

Time Complexity: Meeting Rooms Problem Minimum Rooms Required

O(n log n)

Understanding Time Complexity

We want to find out how the time needed to find the minimum number of meeting rooms changes as the number of meetings grows.

Specifically, how does the code handle more meetings efficiently?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

from heapq import heappush, heappop

def minMeetingRooms(intervals):
    intervals.sort(key=lambda x: x[0])
    heap = []
    for interval in intervals:
        if heap and heap[0] <= interval[0]:
            heappop(heap)
        heappush(heap, interval[1])
    return len(heap)
    

This code finds the minimum number of meeting rooms needed by sorting meetings and using a heap to track end times.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through all meetings once.
• How many times: Once for each meeting (n times).
• Inside loop: Heap operations (push and sometimes pop) happen for each meeting.

How Execution Grows With Input

As the number of meetings increases, the sorting and heap operations grow too.

Input Size (n)	Approx. Operations
10	About 10 log 10 + 10 operations
100	About 100 log 100 + 100 operations
1000	About 1000 log 1000 + 1000 operations

Pattern observation: The time grows a bit faster than just the number of meetings because of sorting and heap operations, but not as fast as checking every pair.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a little more than linearly as meetings increase, mainly due to sorting and managing the heap.

Common Mistake

[X] Wrong: "The code just loops through meetings once, so it must be O(n)."

[OK] Correct: Sorting the meetings first and using a heap for end times adds extra work that grows with n log n, not just n.

Interview Connect

Understanding this time complexity helps you explain how your solution scales and why using sorting plus a heap is efficient for scheduling problems.

Self-Check

"What if we used a simple list instead of a heap to track end times? How would the time complexity change?"

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