Meeting Rooms Problem Minimum Rooms Required in DSA Rust - 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.

How does the algorithm handle more meetings and what costs increase?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn min_meeting_rooms(intervals: &mut Vec<(i32, i32)>) -> i32 {
    intervals.sort_by_key(|k| k.0);
    let mut heap = std::collections::BinaryHeap::new();
    let mut max_rooms: i32 = 0;
    for interval in intervals.iter() {
        while let Some(std::cmp::Reverse(top)) = heap.peek() {
            if *top > interval.0 {
                break;
            }
            let _ = heap.pop();
        }
        heap.push(std::cmp::Reverse(interval.1));
        if (heap.len() as i32) > max_rooms {
            max_rooms = heap.len() as i32;
        }
    }
    max_rooms
}
    

This code sorts meetings by start time, then uses a heap to track ongoing meetings and find the minimum rooms needed.

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).
• Secondary operation: Heap operations (push and pop) inside the loop.
• How many times: Up to n times, each heap operation costs log n.

How Execution Grows With Input

As the number of meetings increases, sorting takes more time, and each meeting causes heap operations that grow with the number of meetings.

Input Size (n)	Approx. Operations
10	Sorting ~ 10 log 10 = 33, Heap ops ~ 10 log 10 = 33
100	Sorting ~ 100 log 100 = 664, Heap ops ~ 100 log 100 = 664
1000	Sorting ~ 1000 log 1000 = 9966, Heap ops ~ 1000 log 1000 = 9966

Pattern observation: The total operations grow roughly like n times log n, which grows faster than just n but slower than n squared.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a bit faster than the number of meetings, mainly because of sorting and heap operations.

Common Mistake

[X] Wrong: "Since we only loop once through meetings, the time is O(n)."

[OK] Correct: Sorting and heap operations inside the loop add extra work that depends on log n, making the total time more than just O(n).

Interview Connect

Understanding this time complexity helps you explain how your solution scales and why sorting and data structures like heaps matter in scheduling problems.

Self-Check

What if we used a simple list instead of a heap to track meeting 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 maximum number of meetings that can be scheduled in one room

B. The total duration of all meetings combined

C. The minimum number of rooms needed so that no meetings overlap in the same room

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. Step 2: Clarify what is being counted

   We count how many meetings overlap at the same time to know how many rooms are needed.

3. Final Answer:

   The minimum number of rooms needed so that no meetings overlap in the same room -> Option C

4. Quick Check:

   Minimum rooms = max overlapping meetings [OK]

Hint: Count max overlapping intervals to find minimum rooms needed [OK]

Common Mistakes:

• Confusing total meetings with rooms needed
• Thinking about meeting durations instead of overlaps
• Assuming rooms equal to total meetings

2. Which Rust data structure is best suited to efficiently track ongoing meetings when solving the Meeting Rooms problem?

easy

A. LinkedList of meeting intervals

B. HashMap with meeting durations as keys

C. Vec (vector) sorted by start times

D. Binary heap (priority queue) to track earliest ending meeting

Solution

1. Step 1: Identify the need to track earliest ending meeting

   We need to know which meeting ends earliest to free a room for the next meeting.

2. Step 2: Choose data structure for efficient retrieval

   A binary heap (priority queue) efficiently gives the earliest ending meeting in O(log n) time.

3. Final Answer:

   Binary heap (priority queue) to track earliest ending meeting -> Option D

4. Quick Check:

   Priority queue tracks earliest end time [OK]

Hint: Use a min-heap to track earliest meeting end times [OK]

Common Mistakes:

• Using HashMap which doesn't keep order
• Using LinkedList which is inefficient for min retrieval
• Using Vec without sorting or heap structure

3. Given the following Rust code snippet, what is the output?

let intervals = vec![(0, 30), (5, 10), (15, 20)];
// Function returns minimum rooms required
let result = min_meeting_rooms(intervals);
println!("{}", result);

fn min_meeting_rooms(mut intervals: Vec<(i32, i32)>) -> i32 {
    intervals.sort_by_key(|k| k.0);
    let mut heap = std::collections::BinaryHeap::new();
    for interval in intervals {
        if let Some(&Reverse(end)) = heap.peek() {
            if interval.0 >= end {
                heap.pop();
            }
        }
        heap.push(Reverse(interval.1));
    }
    heap.len() as i32
}

medium

A. 1

B. 2

C. 3

D. 0

Solution

1. Step 1: Sort intervals by start time

   Sorted intervals: (0,30), (5,10), (15,20)

2. Step 2: Use min-heap to track end times and count rooms

   First meeting (0,30): push 30
   Second meeting (5,10): 5 < 30, push 10 (rooms=2)
   Third meeting (15,20): 15 >= 10, pop 10, push 20 (rooms=2)

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Max overlapping meetings = 2 [OK]

Hint: Sort by start, use min-heap for end times to count rooms [OK]

Common Mistakes:

• Not sorting intervals before processing
• Not popping from heap when room frees
• Counting total meetings instead of overlaps

4. The following Rust code for minimum meeting rooms has a bug. What is the bug?

fn min_meeting_rooms(mut intervals: Vec<(i32, i32)>) -> i32 {
    intervals.sort_by_key(|k| k.1);
    let mut heap = std::collections::BinaryHeap::new();
    for interval in intervals {
        if let Some(&Reverse(end)) = heap.peek() {
            if interval.0 >= end {
                heap.pop();
            }
        }
        heap.push(Reverse(interval.1));
    }
    heap.len() as i32
}

medium

A. Sorting intervals by end time instead of start time causes incorrect room count

B. Using BinaryHeap instead of Vec causes runtime error

C. Not pushing start times into the heap causes logic failure

D. Missing return statement at the end

Solution

1. Step 1: Check sorting criteria

   The code sorts intervals by end time, but we must sort by start time to process meetings in order.

2. Step 2: Understand impact on heap logic

   Sorting by end time breaks the logic of comparing current start with earliest end, leading to wrong room count.

3. Final Answer:

   Sorting intervals by end time instead of start time causes incorrect room count -> Option A

4. Quick Check:

   Sort by start time, not end time [OK]

Hint: Always sort intervals by start time before processing [OK]

Common Mistakes:

• Sorting by end time instead of start time
• Confusing heap usage with wrong keys
• Ignoring heap pop condition

5. You have meetings: [(1,4), (2,5), (7,9), (3,6), (10,11)]. Using the minimum rooms required approach, what is the minimum number of rooms needed?

hard

A. 3

B. 5

C. 4

D. 2

Solution

1. Step 1: Sort intervals by start time

   Sorted: (1,4), (2,5), (3,6), (7,9), (10,11)

2. Step 2: Track overlapping meetings with min-heap

   Push end 4 (room 1)
   Next start 2 < 4, push 5 (room 2)
   Next start 3 < 4, push 6 (room 3)
   Next start 7 >= 4, pop 4, pop 5, pop 6, push 9 (room 1)
   Next start 10 >= 9, pop 9, push 11 (room 1)
   Max rooms used simultaneously = 3

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Max overlaps at once = 3 rooms [OK]

Hint: Count max overlaps after sorting by start time [OK]

Common Mistakes:

• Not sorting intervals before processing
• Counting total meetings instead of overlaps
• Not popping freed rooms from heap