Meeting Rooms II - Practice Coding Problems

Meeting Rooms II - Problem

Array Two Pointers Greedy Sorting Heap (Priority Queue) Prefix Sum Medium

You're managing a busy conference center and need to determine the minimum number of conference rooms required to accommodate all scheduled meetings without any conflicts.

Given an array of meeting time intervals intervals where intervals[i] = [start_i, end_i], your task is to find the minimum number of conference rooms needed so that no two meetings overlap in the same room.

Key Points:

Meetings with the same start and end time as another meeting's end and start time do not overlap
A meeting that ends at time t and another that starts at time t can use the same room
You need to return the minimum number of rooms required

Example: If you have meetings [[0,30],[5,10],[15,20]], you need 2 rooms because the first meeting (0-30) overlaps with both other meetings.

Input & Output

example_1.py — Basic Overlap

$ Input: intervals = [[0,30],[5,10],[15,20]]

› Output: 2

💡 Note: Meeting [0,30] overlaps with both [5,10] and [15,20], but [5,10] and [15,20] don't overlap with each other. So we need 2 rooms: one for [0,30] and another that can be shared by [5,10] and [15,20].

example_2.py — Adjacent Meetings

$ Input: intervals = [[7,10],[2,4]]

› Output: 1

💡 Note: These meetings don't overlap - one ends at 4 and the other starts at 7, so they can share the same room sequentially.

example_3.py — All Overlapping

$ Input: intervals = [[9,10],[4,9],[4,17]]

› Output: 2

💡 Note: [4,9] and [9,10] don't overlap (first ends when second starts), but [4,17] overlaps with both. Maximum concurrent meetings is 2, so we need 2 rooms.

Visualization

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Understanding the Visualization

Receive booking requests

You get meeting requests with start and end times, similar to hotel check-in/check-out

Sort by arrival time

Process meetings chronologically by start time, like handling hotel guests as they arrive

Track room availability

Use a system (heap) to efficiently track when each occupied room becomes free

Optimize room usage

Always reuse the earliest available room to minimize total rooms needed

Key Takeaway

🎯 Key Insight: The minimum number of conference rooms needed equals the maximum number of concurrent meetings at any point in time. Using a min-heap to efficiently track room availability gives us the optimal O(n log n) solution.

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n meetings, we might check against all previously placed meetings in worst case

⚠ Quadratic Growth

Space Complexity

O(n)

We store all meetings distributed across different rooms

⚡ Linearithmic Space

Constraints

1 ≤ intervals.length ≤ 10⁴
0 ≤ start_i < end_i ≤ 10⁶
All meeting times are non-negative integers

Asked in

G Google 45 a Amazon 38 ⊞ Microsoft 32 f Meta 28 Apple 22

The optimal solution uses a min-heap (priority queue) approach. Sort meetings by start time, then use a heap to track when rooms become available. For each meeting, free up rooms that have ended, then allocate the current meeting to a room. The heap size represents the number of rooms needed at any time. Time complexity: O(n log n), Space complexity: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Check All Overlaps)	O(n²)	O(n)	Check every meeting against every other meeting to build room assignments
Two Pointers (Separate Start/End Times)	O(n log n)	O(n)	Separate start and end times, then use two pointers to track concurrent meetings
Min-Heap (Priority Queue) - Optimal	O(n log n)	O(n)	Sort meetings by start time, use min-heap to track room availability efficiently

Brute Force (Check All Overlaps) — Algorithm Steps

Initialize an empty list of rooms (each room contains list of meetings)
For each meeting, try to place it in existing rooms
Check if current meeting overlaps with any meeting in the current room
If no overlap found, add meeting to that room
If overlaps with all existing rooms, create a new room
Return the total number of rooms created

Visualization

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Step-by-Step Walkthrough

Start with empty rooms

Initialize with no rooms allocated

Place first meeting

Create first room and assign first meeting

Check overlaps

For each new meeting, check if it overlaps with meetings in existing rooms

Assign or create

If no overlap, assign to existing room; otherwise create new room

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

bool overlaps(int* meeting1, int* meeting2) {
    return meeting1[0] < meeting2[1] && meeting2[0] < meeting1[1];
}

int minMeetingRooms(int** intervals, int intervalsSize, int* intervalsColSize) {
    if (intervalsSize == 0) return 0;
    
    // Simple approach: for each meeting, count max overlaps
    int maxRooms = 0;
    
    for (int i = 0; i < intervalsSize; i++) {
        int currentRooms = 1;
        
        for (int j = 0; j < intervalsSize; j++) {
            if (i != j && overlaps(intervals[i], intervals[j])) {
                currentRooms++;
            }
        }
        
        if (currentRooms > maxRooms) {
            maxRooms = currentRooms;
        }
    }
    
    return maxRooms;
}

int main() {
    // Input parsing and solution call
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n meetings, we might check against all previously placed meetings in worst case

⚠ Quadratic Growth

Space Complexity

O(n)

We store all meetings distributed across different rooms

⚡ Linearithmic Space

Constraints

1 ≤ intervals.length ≤ 10⁴
0 ≤ start_i < end_i ≤ 10⁶
All meeting times are non-negative integers

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Check All Overlaps) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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