Meeting Rooms Problem Minimum Rooms Required in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the intervals by their start time.

DSA Rust

intervals.sort_by_key(|interval| interval.[1]);

Drag options to blanks, or click blank then click option'

Alen

Bend

Cstart

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by end time instead of start time.

Using incorrect key like length or index.

We sort intervals by their start time to process meetings in chronological order.

2fill in blank

medium

Complete the code to create a min-heap to track meeting end times.

DSA Rust

let mut heap = std::collections::BinaryHeap::new();
heap.push(std::cmp::Reverse(intervals[0].[1]));

Drag options to blanks, or click blank then click option'

Astart

B0

Clen

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Pushing start time instead of end time.

Pushing length or index instead of time.

We push the end time of the first meeting to track when a room becomes free.

3fill in blank

hard

Fix the error in the condition to check if the current meeting can reuse a room.

DSA Rust

if intervals[i].start [1] heap.peek().unwrap().0 {
    heap.pop();
}

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A>=

B<

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '<=' which is incorrect for this logic.

Using '==' only which misses cases where start is after end.

We check if the current meeting starts after or exactly when the earliest meeting ends to reuse the room.

4fill in blank

hard

Fill all three blanks to correctly update the heap with the current meeting's end time and iterate over intervals.

DSA Rust

for [1] in 1..intervals.len() {
    if intervals[[2]].start >= heap.peek().unwrap().0 {
        heap.pop();
    }
    heap.push(std::cmp::Reverse(intervals[[3]].end));
}

Drag options to blanks, or click blank then click option'

Ai

Bj

Ck

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for loop and indexing causing errors.

Using undefined variables.

We use 'i' as the loop variable consistently to access intervals and update the heap.

5fill in blank

hard

Fill all three blanks to return the minimum number of rooms required after processing all intervals.

DSA Rust

fn min_meeting_rooms(intervals: &mut Vec<Interval>) -> usize {
    intervals.sort_by_key(|interval| interval.start);
    let mut heap = std::collections::BinaryHeap::new();
    heap.push(std::cmp::Reverse(intervals[0].end));
    for [1] in 1..intervals.len() {
        if intervals[[2]].start >= heap.peek().unwrap().0 {
            heap.pop();
        }
        heap.push(std::cmp::Reverse(intervals[[3]].end));
    }
    heap.len()
}

Drag options to blanks, or click blank then click option'

Ai

Bj

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables causing indexing errors.

Returning wrong value instead of heap length.

Using 'i' consistently for loop and indexing ensures correct processing and final count of rooms.

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