Meeting Rooms Problem Minimum Rooms Required in DSA Rust - Build from Scratch

Meeting Rooms Problem Minimum Rooms Required

📖 Scenario: You are organizing a conference with multiple meetings scheduled at different times. You want to find out the minimum number of meeting rooms needed so that no meetings overlap in the same room.

🎯 Goal: Build a Rust program that calculates the minimum number of meeting rooms required given a list of meeting time intervals.

📋 What You'll Learn

Create a vector of tuples representing meeting intervals with exact values

Create a variable to hold the count of meeting rooms needed

Implement the logic to find the minimum number of meeting rooms required

Print the final minimum number of meeting rooms

💡 Why This Matters

🌍 Real World

Scheduling meetings or events in rooms without conflicts is a common problem in offices, schools, and conference centers.

💼 Career

Understanding interval scheduling and resource allocation is important for software engineers working on calendar apps, booking systems, and event management software.

1

Create the meeting intervals data

Create a vector called meetings with these exact tuples representing meeting start and end times: (0, 30), (5, 10), (15, 20)

DSA Rust

# Your code here

Hint

Use let meetings = vec![(0, 30), (5, 10), (15, 20)]; to create the vector.

2

Create a variable to track the minimum rooms needed

Create a mutable variable called min_rooms and set it to 0 to track the minimum number of meeting rooms needed

DSA Rust

fn main() {
    let meetings = vec![(0, 30), (5, 10), (15, 20)];
    // Create min_rooms variable here
    # Your code here
}

Hint

Use let mut min_rooms = 0; to create the variable.

3

Implement the logic to find minimum meeting rooms required

Sort the meetings vector by start time, create two vectors start_times and end_times containing start and end times respectively, then use two indices i and j to iterate and update min_rooms to find the minimum number of rooms required

DSA Rust

fn main() {
    let mut meetings = vec![(0, 30), (5, 10), (15, 20)];
    let mut min_rooms = 0;
    // Implement sorting and logic to find min_rooms
    # Your code here
}

Hint

Sort meetings by start time, then use two pointers to compare start and end times to count rooms.

4

Print the minimum number of meeting rooms required

Print the value of min_rooms using println!("{}", min_rooms);

DSA Rust

fn main() {
    let mut meetings = vec![(0, 30), (5, 10), (15, 20)];
    let mut min_rooms = 0;

    meetings.sort_by_key(|k| k.0);

    let start_times: Vec = meetings.iter().map(|x| x.0).collect();
    let end_times: Vec = meetings.iter().map(|x| x.1).collect();

    let mut i = 0;
    let mut j = 0;
    let mut rooms_in_use = 0;

    while i  min_rooms {
                min_rooms = rooms_in_use;
            }
            i += 1;
        } else {
            rooms_in_use -= 1;
            j += 1;
        }
    }
    // Print min_rooms here
    # Your code here
}

Hint

Use println!("{}", min_rooms); to print the result.

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