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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Typescript

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

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

A-

Ba.start - b.start

Cstart - end

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect properties like 'start - end' which is invalid.

Using just '-' without properties.

Using '.start - b.start' which is syntax error.

We sort intervals by their start time using a.start - b.start to order them from earliest to latest.

2fill in blank

medium

Complete the code to check if the current meeting starts after the earliest ending meeting.

DSA Typescript

if (intervals[i].start >= [1]) {

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

AminHeap.peek()

BminHeap[0]

Cheap[0]

Dheap.peek()

Attempts:

3 left

💡 Hint

Common Mistakes

Using array indexing like heap[0] when min heap is a class.

Using minHeap[0] which may not be valid for heap structure.

We use minHeap.peek() to get the earliest ending meeting's end time without removing it.

3fill in blank

hard

Fix the error in the code to add the current meeting's end time to the min heap.

DSA Typescript

minHeap.[1](intervals[i].end);

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

Ainsert

Badd

Cpush

Denqueue

Attempts:

3 left

💡 Hint

Common Mistakes

Using push() which is for arrays, not heaps.

Using add() or enqueue() which may not exist in this heap class.

The min heap uses insert() method to add new elements while maintaining heap property.

4fill in blank

hard

Fill both blanks to remove the earliest ending meeting and add the current meeting's end time.

DSA Typescript

if (intervals[i].start >= minHeap.peek()) {
  minHeap.[1]();
}
minHeap.[2](intervals[i].end);

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

AextractMin

Binsert

Cremove

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() which may not be defined for min heap.

Using remove() which is ambiguous.

Using insert() before removing the earliest meeting.

extractMin() removes the smallest element, and insert() adds the current meeting's end time.

5fill in blank

hard

Fill the blank to return the minimum number of meeting rooms needed.

DSA Typescript

return minHeap.[1];

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

Acount

Blength

Csize

Dcapacity

Attempts:

3 left

💡 Hint

Common Mistakes

Using length which is for arrays, not heap.

Using count or capacity which may not exist.

The size property of the min heap gives the number of rooms currently used.

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 duration of a single meeting

B. The minimum number of meeting rooms needed to hold all meetings without overlap

C. The total number of meetings scheduled

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 meetings overlap in the same room.

2. Step 2: Clarify what is NOT asked

   It is not about meeting durations, counts, or earliest times, but about room allocation to avoid conflicts.

3. Final Answer:

   The minimum number of meeting rooms needed to hold all meetings without overlap -> Option B

4. Quick Check:

   Minimum rooms = no overlapping meetings [OK]

Hint: Count overlaps to find minimum rooms needed [OK]

Common Mistakes:

• Confusing total meetings with rooms needed
• Thinking about meeting length instead of overlaps
• Ignoring overlapping intervals

2. Which TypeScript code snippet correctly sorts an array of meeting intervals by their start time?

easy

A. intervals.sort((a, b) => a[0] + b[0]);

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

C. intervals.sort((a, b) => a[1] - b[0]);

D. intervals.sort((a, b) => a[0] - b[0]);

Solution

1. Step 1: Identify sorting by start time

   We want to sort intervals by their start time, which is the first element in each interval array.

2. Step 2: Check the sorting function

   The correct comparator subtracts start times: a[0] - b[0]. This sorts ascending by start time.

3. Final Answer:

   intervals.sort((a, b) => a[0] - b[0]); -> Option D

4. Quick Check:

   Sort by start time = a[0] - b[0] [OK]

Hint: Sort intervals by start time using a[0] - b[0] [OK]

Common Mistakes:

• Sorting by end time instead of start time
• Using addition instead of subtraction in comparator
• Mixing start and end indices in comparator

3. Given the following TypeScript code, what is the output?

const intervals = [[0, 30], [5, 10], [15, 20]];

function minMeetingRooms(intervals: number[][]): number {
  intervals.sort((a, b) => a[0] - b[0]);
  const heap: number[] = [];

  for (const interval of intervals) {
    if (heap.length && heap[0] <= interval[0]) {
      heap.shift();
    }
    heap.push(interval[1]);
    heap.sort((a, b) => a - b);
  }

  return heap.length;
}

console.log(minMeetingRooms(intervals));

medium

A. 2

B. 3

C. 1

D. 0

Solution

1. Step 1: Sort intervals by start time

   Sorted intervals: [[0,30], [5,10], [15,20]] (already sorted)

2. Step 2: Simulate heap to track end times

   First interval: heap = [30] Second interval starts at 5, heap[0]=30 > 5, so no removal, heap = [30,10], sorted to [10,30] Third interval starts at 15, heap[0]=10 <= 15, remove 10, heap = [30], then add 20, heap = [20,30]

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Max overlapping meetings = 2 rooms needed [OK]

Hint: Track earliest end time to reuse rooms [OK]

Common Mistakes:

• Not sorting intervals before processing
• Not removing ended meetings from heap
• Assuming heap length equals total meetings

4. The following TypeScript code is intended to find the minimum number of meeting rooms needed. What is the bug?

function minMeetingRooms(intervals: number[][]): number {
  intervals.sort((a, b) => a[1] - b[1]);
  const heap: number[] = [];

  for (const interval of intervals) {
    if (heap.length && heap[0] <= interval[0]) {
      heap.shift();
    }
    heap.push(interval[1]);
    heap.sort((a, b) => a - b);
  }

  return heap.length;
}

medium

A. Intervals are sorted by end time instead of start time

B. Heap is not sorted after push

C. The condition to remove from heap is incorrect

D. The function returns the wrong value

Solution

1. Step 1: Check sorting criteria

   The intervals are sorted by end time (a[1] - b[1]) but should be sorted by start time to process meetings in order.

2. Step 2: Understand impact of wrong sorting

   Sorting by end time breaks the logic of checking overlaps because meetings are not processed in chronological start order.

3. Final Answer:

   Intervals are sorted by end time instead of start time -> Option A

4. Quick Check:

   Sort by start time, not end time [OK]

Hint: Always sort intervals by start time first [OK]

Common Mistakes:

• Sorting by end time instead of start time
• Not sorting heap after insertion
• Incorrect condition to remove ended meetings

5. You have the following meeting intervals: [[1,4], [2,5], [7,9], [3,6], [10,11]]. What is the minimum number of meeting rooms required to schedule all meetings without overlap?

hard

A. 4

B. 2

C. 3

D. 1

Solution

1. Step 1: Sort intervals by start time

   Sorted intervals: [[1,4], [2,5], [3,6], [7,9], [10,11]]

2. Step 2: Track overlapping meetings

   At time 1: room count = 1 At time 2: overlaps with [1,4], rooms = 2 At time 3: overlaps with [1,4] and [2,5], rooms = 3 At time 7: previous meetings ended, rooms = 1 At time 10: rooms = 1

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Max overlaps = 3 rooms needed [OK]

Hint: Count max overlaps at any time for rooms needed [OK]

Common Mistakes:

• Not sorting intervals before counting overlaps
• Underestimating overlaps by ignoring simultaneous meetings
• Assuming one room fits all meetings