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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

intervals.sort((a, b) => a[1]b);

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

A.start - b.start

Bstart - end

C-

D[0] - b[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect properties like 'start' when intervals are arrays.

Not subtracting correctly causing wrong sort order.

We sort intervals by their start time which is at index 0, so we subtract b[0] from a[0].

2fill in blank

medium

Complete the code to push the end time of the first meeting into the heap.

DSA Javascript

const heap = [];
heap.push(intervals[0][1]);

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

Aend

B.end

C[1]

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 0 which is the start time instead of end time.

Trying to access a property 'end' which does not exist.

We push the end time of the first meeting which is at index 1 in the interval array.

3fill in blank

hard

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

DSA Javascript

if (intervals[i][0] [1] heap[0]) {
  heap.shift();
}

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

A<=

B>=

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '<=' which reverses the logic.

Using '>' which misses the equal case.

If the current meeting starts at or after the earliest ending meeting, we can reuse that room.

4fill in blank

hard

Fill both blanks to insert the current meeting's end time and keep the heap sorted.

DSA Javascript

heap.push(intervals[i][1]);
heap.sort((a, b) => a [2] b);

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

A[1]

B-

C+

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Pushing start time instead of end time.

Sorting with '+' which does not sort correctly.

We push the end time at index 1 and sort the heap by ascending end times using subtraction.

5fill in blank

hard

Fill all three blanks to return the minimum number of rooms needed.

DSA Javascript

function minMeetingRooms(intervals) {
  if (intervals.length === 0) return 0;
  intervals.sort((a, b) => a[0] - b[0]);
  const heap = [];
  heap.push(intervals[0][1]);
  for (let i = 1; i < intervals.length; i++) {
    if (intervals[i][0] [2] heap[0]) {
      heap.shift();
    }
    heap.push(intervals[i][3]);
    heap.sort((a, b) => a - b);
  }
  return heap.length;
}

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

A[1]

B>=

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using start times instead of end times for heap operations.

Incorrect comparison operator in the if condition.

We push end times at index 1, check if current start time is >= earliest end time, and push current end time again at index 1.

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 minimum number of meeting rooms needed so no meetings overlap

B. The maximum length of a single meeting

C. The total number of meetings scheduled

D. The earliest meeting start time

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum number of rooms required so that no two meetings overlap in the same room.

2. Step 2: Clarify what is being counted

   We count how many meetings happen at the same time to find the minimum rooms needed.

3. Final Answer:

   The minimum number of meeting rooms needed so no meetings overlap -> Option A

4. Quick Check:

   Minimum rooms = max overlapping meetings [OK]

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

Common Mistakes:

• Confusing total meetings with rooms needed
• Ignoring overlapping times
• Thinking earliest start time is the answer

2. Which of the following JavaScript code snippets correctly sorts an array of meeting intervals by their start time? Assume intervals is an array of objects with 'start' and 'end' properties.

easy

A. intervals.sort((a, b) => b.end - a.end);

B. intervals.sort((a, b) => b.start - a.start);

C. intervals.sort((a, b) => a.end - b.end);

D. intervals.sort((a, b) => a.start - b.start);

Solution

1. Step 1: Identify sorting by start time

   We want to sort intervals by their start property in ascending order.

2. Step 2: Check the sort function

   The comparator (a, b) => a.start - b.start sorts ascending by start time.

3. Final Answer:

   intervals.sort((a, b) => a.start - b.start); -> Option D

4. Quick Check:

   Ascending start time sort = intervals.sort((a, b) => a.start - b.start); [OK]

Hint: Sort intervals by start ascending: (a,b) => a.start - b.start [OK]

Common Mistakes:

• Sorting by end time instead of start
• Sorting descending when ascending needed
• Using incorrect comparator syntax

3. What is the minimum number of meeting rooms required for the following intervals?
const intervals = [{start: 0, end: 30}, {start: 5, end: 10}, {start: 15, end: 20}];

medium

A. 1

B. 2

C. 3

D. 4

Solution

1. Step 1: Analyze overlapping intervals

   Meetings: [0-30], [5-10], [15-20]. The first overlaps with the second and third.

2. Step 2: Count max overlaps

   At time 5-10, two meetings overlap (0-30 and 5-10). At 15-20, two meetings overlap (0-30 and 15-20). Max overlap is 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Max overlapping meetings = 2 rooms needed [OK]

Hint: Find max overlaps by checking intervals start and end times [OK]

Common Mistakes:

• Counting total meetings instead of overlaps
• Assuming 3 rooms because 3 meetings exist
• Ignoring overlap intervals

4. The following code attempts to find the minimum number of meeting rooms needed but has a bug. What is the bug?

function minMeetingRooms(intervals) {
  intervals.sort((a, b) => a.start - b.start);
  let rooms = [];
  for (let meeting of intervals) {
    if (rooms.length === 0 || meeting.start < rooms[rooms.length - 1]) {
      rooms.push(meeting.end);
    } else {
      rooms[rooms.length - 1] = meeting.end;
    }
  }
  return rooms.length;
}

medium

A. It returns the wrong value (should return rooms array)

B. It sorts intervals by end time instead of start time

C. It compares meeting.start with a number incorrectly (rooms array stores end times, but comparison is wrong)

D. It uses push instead of unshift on rooms array

Solution

1. Step 1: Understand rooms array usage

   Rooms array stores end times of meetings currently occupying rooms.

2. Step 2: Check comparison logic

   Comparing meeting.start with rooms[rooms.length - 1] (last end time) is incorrect because rooms should be a min-heap or sorted structure, not just last element.

3. Final Answer:

   It compares meeting.start with a number incorrectly (rooms array stores end times, but comparison is wrong) -> Option C

4. Quick Check:

   Incorrect comparison of start with last end time causes bug [OK]

Hint: Use a min-heap or sorted structure for end times, not last element [OK]

Common Mistakes:

• Comparing start with last end time without sorting
• Not using a priority queue for room end times
• Returning wrong variable

5. Given the intervals [[1,4],[2,5],[7,9],[6,10],[8,11]], what is the minimum number of meeting rooms required? Explain your reasoning.

hard

A. 3

B. 2

C. 4

D. 5

Solution

1. Step 1: Sort intervals by start time

   Sorted intervals: [1,4], [2,5], [6,10], [7,9], [8,11]

2. Step 2: Track overlapping meetings

   Between 1-5, two meetings overlap ([1,4] and [2,5]). Between 6-11, three meetings overlap ([6,10], [7,9], [8,11]). Max overlap is 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Max overlapping meetings = 3 rooms needed [OK]

Hint: Find max overlaps by checking intervals start and end times carefully [OK]

Common Mistakes:

• Not sorting intervals before checking overlaps
• Counting total meetings instead of max overlaps
• Missing overlaps in later intervals