Job Scheduling with Deadlines in DSA Typescript - Build from Scratch

Job Scheduling with Deadlines

📖 Scenario: You are managing a small workshop where several jobs need to be completed. Each job has a deadline and a profit if finished before or on that deadline. You want to schedule jobs to maximize your total profit.

🎯 Goal: Build a simple program to schedule jobs with deadlines to maximize profit. You will create the job list, set a maximum deadline, select jobs to maximize profit, and then print the scheduled jobs.

📋 What You'll Learn

Create an array of jobs with exact id, deadline, and profit

Create a variable for the maximum deadline

Implement the job scheduling logic to select jobs maximizing profit

Print the scheduled jobs in order of their scheduled time slots

💡 Why This Matters

🌍 Real World

Job scheduling is used in factories, computer task management, and project planning to maximize efficiency and profit.

💼 Career

Understanding job scheduling helps in roles like software development, operations management, and systems engineering where task prioritization is key.

1

Create the job list

Create an array called jobs with these exact objects: {id: 'a', deadline: 2, profit: 100}, {id: 'b', deadline: 1, profit: 19}, {id: 'c', deadline: 2, profit: 27}, {id: 'd', deadline: 1, profit: 25}, {id: 'e', deadline: 3, profit: 15}

DSA Typescript

// Create the jobs array with given job objects
// Your code here

Hint

Use const jobs = [ ... ] with the exact job objects inside.

2

Set the maximum deadline

Create a variable called maxDeadline and set it to 3, which is the maximum deadline among the jobs.

DSA Typescript

const jobs = [
  {id: 'a', deadline: 2, profit: 100},
  {id: 'b', deadline: 1, profit: 19},
  {id: 'c', deadline: 2, profit: 27},
  {id: 'd', deadline: 1, profit: 25},
  {id: 'e', deadline: 3, profit: 15}
];
// Create maxDeadline variable and set it to 3
// Your code here

Hint

Use const maxDeadline = 3; to set the maximum deadline.

3

Schedule jobs to maximize profit

Sort the jobs array by profit in descending order. Create an array called result of length maxDeadline filled with null. For each job in sorted order, find a free slot from its deadline backwards and assign the job's id to that slot in result if free.

DSA Typescript

const jobs = [
  {id: 'a', deadline: 2, profit: 100},
  {id: 'b', deadline: 1, profit: 19},
  {id: 'c', deadline: 2, profit: 27},
  {id: 'd', deadline: 1, profit: 25},
  {id: 'e', deadline: 3, profit: 15}
];

const maxDeadline = 3;

// Sort jobs by profit descending and schedule them
// Your code here

Hint

Sort jobs by profit descending. Use a loop to assign each job to the latest free slot before its deadline.

4

Print the scheduled jobs

Print the result array showing scheduled job ids separated by -> . For empty slots, print null.

DSA Typescript

const jobs = [
  {id: 'a', deadline: 2, profit: 100},
  {id: 'b', deadline: 1, profit: 19},
  {id: 'c', deadline: 2, profit: 27},
  {id: 'd', deadline: 1, profit: 25},
  {id: 'e', deadline: 3, profit: 15}
];

const maxDeadline = 3;

jobs.sort((a, b) => b.profit - a.profit);

const result: (string | null)[] = new Array(maxDeadline).fill(null);

for (const job of jobs) {
  for (let slot = Math.min(job.deadline, maxDeadline) - 1; slot >= 0; slot--) {
    if (result[slot] === null) {
      result[slot] = job.id;
      break;
    }
  }
}

// Print the scheduled jobs separated by ' -> '
// Your code here

Hint

Use console.log(result.map(x => x === null ? 'null' : x).join(' -> ')) to print the scheduled jobs.

Practice

1. What is the main goal of the Job Scheduling with Deadlines problem?

easy

A. Maximize the number of jobs without considering profit

B. Minimize the number of jobs scheduled

C. Schedule jobs in the order they appear without considering deadlines

D. Maximize the total profit by scheduling jobs before their deadlines

Solution

1. Step 1: Understand the problem goal

   The problem asks to schedule jobs so that total profit is maximized while respecting deadlines.

2. Step 2: Identify the objective

   Scheduling jobs before their deadlines to get maximum profit is the main goal.

3. Final Answer:

   Maximize the total profit by scheduling jobs before their deadlines -> Option D

4. Quick Check:

   Goal = Maximize profit with deadlines [OK]

Hint: Focus on maximizing profit within deadlines [OK]

Common Mistakes:

• Ignoring deadlines and scheduling all jobs
• Trying to minimize jobs instead of maximizing profit
• Scheduling jobs in input order without sorting

2. Which of the following TypeScript code snippets correctly declares a job interface with id, deadline, and profit properties?

easy

A. type Job = { id: string; deadline: string; profit: string }

B. class Job { id; deadline; profit; }

C. interface Job { id: number; deadline: number; profit: number; }

D. interface Job { id; deadline; profit; }

Solution

1. Step 1: Check correct TypeScript interface syntax

   interface Job { id: number; deadline: number; profit: number; } correctly declares an interface with typed properties.

2. Step 2: Identify incorrect options

   type Job = { id: string; deadline: string; profit: string } uses strings instead of numbers, class Job { id; deadline; profit; } is a class without types, interface Job { id; deadline; profit; } misses types.

3. Final Answer:

   interface Job { id: number; deadline: number; profit: number; } -> Option C

4. Quick Check:

   Correct interface syntax = interface Job { id: number; deadline: number; profit: number; } [OK]

Hint: Use interface with typed properties for job data [OK]

Common Mistakes:

• Using class without types for data structure
• Omitting types in interface properties
• Using wrong types like string for numeric fields

3. What will be the output of the following TypeScript code that schedules jobs to maximize profit?

interface Job {
  id: string;
  deadline: number;
  profit: number;
}

const jobs: Job[] = [
  { id: 'a', deadline: 2, profit: 100 },
  { id: 'b', deadline: 1, profit: 19 },
  { id: 'c', deadline: 2, profit: 27 },
  { id: 'd', deadline: 1, profit: 25 },
  { id: 'e', deadline: 3, profit: 15 }
];

function scheduleJobs(jobs: Job[]): string[] {
  jobs.sort((a, b) => b.profit - a.profit);
  const result: (string | null)[] = Array(jobs.length).fill(null);
  for (const job of jobs) {
    for (let j = Math.min(jobs.length, job.deadline) - 1; j >= 0; j--) {
      if (result[j] === null) {
        result[j] = job.id;
        break;
      }
    }
  }
  return result.filter(id => id !== null) as string[];
}

console.log(scheduleJobs(jobs));

medium

A. ['a', 'd', 'e']

B. ['a', 'c', 'e']

C. ['b', 'd', 'e']

D. ['a', 'b', 'c']

Solution

1. Step 1: Sort jobs by profit descending

   Sorted jobs: a(100), c(27), d(25), b(19), e(15).

2. Step 2: Schedule jobs in latest available slot before deadline

   Job 'a' scheduled at slot 1 (deadline 2), 'c' at slot 0 (deadline 2), 'd' cannot fit before deadline 1 (slot 0 taken), 'b' cannot fit, 'e' scheduled at slot 2 (deadline 3).

3. Final Answer:

   ['a', 'c', 'e'] -> Option B

4. Quick Check:

   Scheduled jobs = ['a', 'c', 'e'] [OK]

Hint: Sort by profit, assign latest free slot before deadline [OK]

Common Mistakes:

• Scheduling jobs in input order without sorting
• Assigning earliest slot instead of latest
• Ignoring deadline constraints

4. Identify the bug in the following TypeScript function for job scheduling and select the correct fix.

function scheduleJobs(jobs: Job[]): string[] {
  jobs.sort((a, b) => a.profit - b.profit);
  const result: (string | null)[] = Array(jobs.length).fill(null);
  for (const job of jobs) {
    for (let j = job.deadline - 1; j >= 0; j--) {
      if (result[j] === null) {
        result[j] = job.id;
        break;
      }
    }
  }
  return result.filter(id => id !== null) as string[];
}

medium

A. Sort jobs by profit descending instead of ascending

B. Initialize result array with zeros instead of null

C. Loop j from 0 to job.deadline - 1 instead of reverse

D. Remove the inner loop and assign jobs directly

Solution

1. Step 1: Analyze sorting order

   Jobs are sorted by profit ascending, which schedules low profit jobs first, reducing total profit.

2. Step 2: Correct sorting order

   Sorting by profit descending ensures high profit jobs are scheduled first for maximum profit.

3. Final Answer:

   Sort jobs by profit descending instead of ascending -> Option A

4. Quick Check:

   Profit descending sort fixes scheduling [OK]

Hint: Sort jobs by descending profit to fix scheduling [OK]

Common Mistakes:

• Sorting by ascending profit reduces total profit
• Changing loop direction incorrectly
• Using zero instead of null for empty slots

5. Given the following jobs with deadlines and profits, what is the maximum total profit achievable by scheduling jobs before their deadlines? Jobs: - Job1: deadline = 4, profit = 70 - Job2: deadline = 1, profit = 80 - Job3: deadline = 1, profit = 30 - Job4: deadline = 2, profit = 100 - Job5: deadline = 3, profit = 40 Select the correct maximum profit.

hard

A. 290

B. 210

C. 240

D. 230

Solution

1. Step 1: Sort jobs by profit descending

   Order: Job4(100), Job2(80), Job1(70), Job5(40), Job3(30).

2. Step 2: Schedule jobs in latest free slot before deadline

   Job4 at slot 1 (deadline 2), Job2 at slot 0 (deadline 1), Job1 at slot 3 (deadline 4), Job5 at slot 2 (deadline 3), Job3 cannot fit (slot 0 taken).

3. Step 3: Calculate total profit

   100 + 80 + 70 + 40 = 290, but slots are 0-based: slots 0,1,2,3 used. Actually, Job2 (deadline 1) at slot 0, Job4 (deadline 2) at slot 1, Job5 (deadline 3) at slot 2, Job1 (deadline 4) at slot 3. Total profit = 100 + 80 + 40 + 70 = 290.

4. Step 4: Re-check options

   Confirm scheduling: slots 0:Job2, 1:Job4, 2:Job5, 3:Job1. All fit, total 290.

5. Final Answer:

   290 -> Option A

6. Quick Check:

   Max profit = 290 [OK]

Hint: Schedule highest profit jobs first in latest free slot [OK]

Common Mistakes:

• Assuming all jobs fit without checking slots
• Adding profits of unscheduled jobs
• Misinterpreting deadlines as exact time slots