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

Meeting Rooms Problem Minimum Rooms Required

📖 Scenario: You are organizing meetings in a company. Each meeting has a start time and an end time. You want to find out the minimum number of meeting rooms needed so that all meetings can happen without overlapping in the same room.

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

📋 What You'll Learn

Create an array of meeting intervals with exact start and end times

Create a variable to keep track of the number of rooms needed

Write logic to find the minimum number of meeting rooms required

Print the minimum number of meeting rooms needed

💡 Why This Matters

🌍 Real World

Scheduling meetings in offices or conference centers to avoid conflicts and ensure enough rooms are available.

💼 Career

This problem is common in software engineering roles involving scheduling, calendar apps, and resource allocation.

1

Create the meeting intervals array

Create an array called meetings with these exact intervals: [1, 4], [2, 5], [7, 9]

DSA Typescript

# Create the array called meetings with the given intervals
// Your code here

Hint

Use a constant array named meetings with tuples for each interval.

2

Create a variable to track rooms needed

Create a variable called roomsNeeded and set it to 0

DSA Typescript

const meetings: [number, number][] = [[1, 4], [2, 5], [7, 9]];
// Create a variable called roomsNeeded and set it to 0
// Your code here

Hint

Use let to create a variable that can change later.

3

Calculate minimum meeting rooms required

Write code to calculate the minimum number of meeting rooms needed using the meetings array and store the result in roomsNeeded

DSA Typescript

const meetings: [number, number][] = [[1, 4], [2, 5], [7, 9]];
let roomsNeeded = 0;
// Calculate minimum rooms needed and assign to roomsNeeded
// Your code here

Hint

Sort start and end times separately. Use two pointers to compare them and count rooms.

4

Print the minimum number of meeting rooms needed

Write a console.log statement to print the value of roomsNeeded

DSA Typescript

const meetings: [number, number][] = [[1, 4], [2, 5], [7, 9]];
let roomsNeeded = 0;

const starts = meetings.map(m => m[0]).sort((a, b) => a - b);
const ends = meetings.map(m => m[1]).sort((a, b) => a - b);

let startPointer = 0;
let endPointer = 0;

while (startPointer < meetings.length) {
  if (starts[startPointer] < ends[endPointer]) {
    roomsNeeded++;
    startPointer++;
  } else {
    endPointer++;
    startPointer++;
  }
}
// Print the roomsNeeded value
// Your code here

Hint

Use console.log(roomsNeeded) to show 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 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