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

Concept Flow - Meeting Rooms Problem Minimum Rooms Required

Sort intervals by start time

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Initialize min-heap for end times

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For each interval

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If min-heap empty or earliest end > current start

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Add new room (push end time)

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Update rooms count

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Else earliest end <= current start

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Reuse room (pop earliest end, push current end)

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Repeat for all intervals

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Return size of min-heap as minimum rooms

Sort meetings by start time, use a min-heap to track earliest ending meeting, allocate or reuse rooms accordingly.

Execution Sample

DSA Python

import heapq
intervals = [[0,30],[5,10],[15,20]]
intervals.sort(key=lambda x: x[0])
rooms = []
for interval in intervals:
    if rooms and rooms[0] <= interval[0]:
        heapq.heappop(rooms)
    heapq.heappush(rooms, interval[1])
print(len(rooms))

Calculate minimum meeting rooms needed for given intervals by sorting and using a min-heap for end times.

Execution Table

Step	Operation	Current Interval	Min-Heap Before	Heap Action	Min-Heap After	Rooms Count	Visual State
1	Sort intervals	[[0,30],[5,10],[15,20]]	[]	Sort by start	[]	0	Intervals sorted by start time
2	Process interval	[0,30]	[]	Push end=30	[30]	1	Rooms: 30
3	Process interval	[5,10]	[30]	Push end=10	[10,30]	2	Rooms: 10 -> 30
4	Process interval	[15,20]	[10,30]	Pop 10, Push 20	[20,30]	2	Rooms: 20 -> 30
5	End	All intervals processed	[20,30]	None	[20,30]	2	Minimum rooms required = 2

💡 All intervals processed, min-heap size 2 means 2 rooms needed

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]]
rooms (min-heap)	[]	[30]	[10,30]	[20,30]	[20,30]
rooms count	0	1	2	2	2

Key Moments - 3 Insights

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

Because the room with the earliest end time is now free and can be reused for the current meeting, so we remove that end time before adding the current meeting's end time. See execution_table row 4.

Why do we use a min-heap to store end times instead of just a list?

A min-heap efficiently gives the earliest ending meeting in O(1) time, which helps decide if a room can be reused quickly. This is shown in the heap actions in execution_table rows 2-4.

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

Because each end time in the heap represents a meeting currently occupying a room. The heap size shows how many rooms are in use simultaneously. See final step in execution_table.

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 [10,30]

C [30]

D [20,30]

💡 Hint

Check the 'Min-Heap Before' column at Step 3 in execution_table.

At which step does the algorithm reuse a room by popping from the min-heap?

A Step 4

B Step 2

C Step 3

D Step 5

💡 Hint

Look for 'Pop' action in the 'Heap Action' column in execution_table.

If the interval [5,10] was instead [35,40], how would the rooms count change after processing it?

A Rooms count would increase to 3

B Rooms count would remain 2

C Rooms count would decrease to 1

D Rooms count would be 0

💡 Hint

Consider that [35,40] starts after earliest end time in min-heap; see variable_tracker and heap reuse logic.

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 from heap (reuse room)
  - Push current end time
- Heap size = minimum rooms needed

Full Transcript

This visualization shows how to find the minimum number of meeting rooms needed for a list of meeting intervals. First, we sort the intervals by their start times. Then, we use a min-heap to keep track of the end times of meetings currently occupying rooms. For each meeting, if the earliest ending meeting ends before or when the current meeting starts, we reuse that room by popping from the heap. Otherwise, we allocate a new room by pushing the current meeting's end time. The size of the min-heap at the end tells us how many rooms are needed. The execution table traces each step, showing the heap state and room count. Key moments clarify why we pop from the heap and why the heap size represents rooms needed. The quiz tests understanding of heap states and room reuse.