Job Scheduling with Deadlines in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Job Scheduling with Deadlines and Profits

What is the output of the following TypeScript code that schedules jobs to maximize profit given deadlines?

DSA Typescript

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

function jobScheduling(jobs: Job[]): string {
  jobs.sort((a, b) => b.profit - a.profit);
  const n = jobs.length;
  const result: (string | null)[] = Array(n).fill(null);
  const slot: boolean[] = Array(n).fill(false);

  for (let i = 0; i < n; i++) {
    for (let j = Math.min(n - 1, jobs[i].deadline - 1); j >= 0; j--) {
      if (!slot[j]) {
        slot[j] = true;
        result[j] = jobs[i].id;
        break;
      }
    }
  }

  return result.filter(x => x !== null).join(' -> ') + ' -> null';
}

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 }
];

console.log(jobScheduling(jobs));

A "a -> c -> e -> null"

B "a -> d -> e -> null"

C "b -> d -> e -> null"

D "a -> b -> c -> null"

💡 Hint

Sort jobs by profit descending and assign each job to the latest available slot before its deadline.

The jobs are sorted by profit: a(100), c(27), d(25), b(19), e(15).

Assign 'a' to slot 1 (deadline 2), 'c' to slot 0 (deadline 2), and 'e' to slot 2 (deadline 3). Slots 0,1,2 correspond to time slots 1,2,3 respectively.

Resulting schedule is "a -> c -> e -> null".

🧠 Conceptual

intermediate

Maximum Number of Jobs Scheduled

Given 5 jobs with deadlines [2, 1, 2, 1, 3] and profits [100, 19, 27, 25, 15], what is the maximum number of jobs that can be scheduled without missing deadlines?

A 2

B 4

C 3

D 5

💡 Hint

Each job takes 1 unit of time. Schedule jobs greedily by profit and assign to latest free slot before deadline.

Maximum slots available are 3 (max deadline).

By scheduling jobs with highest profit first, we can schedule 3 jobs without conflicts.

🔧 Debug

advanced

Identify the Logical Error in Job Scheduling Code

What is the logical error in the following TypeScript code snippet for job scheduling with deadlines? function jobScheduling(jobs: Job[]): string { jobs.sort((a, b) => a.profit - b.profit); const n = jobs.length; const result: (string | null)[] = Array(n).fill(null); const slot: boolean[] = Array(n).fill(false); for (let i = 0; i < n; i++) { for (let j = Math.min(n - 1, jobs[i].deadline - 1); j >= 0; j--) { if (!slot[j]) { slot[j] = true; result[j] = jobs[i].id; break; } } } return result.filter(x => x !== null).join(' -> ') + ' -> null'; }

A Slots array is not initialized correctly, causing index out of range errors.

B The result array should store profits instead of job ids for correct output.

C The inner loop should iterate forwards, not backwards, to assign jobs correctly.

D Jobs are sorted in ascending order of profit instead of descending, causing suboptimal scheduling.

💡 Hint

Check the sorting order of jobs by profit.

The code sorts jobs by ascending profit, which schedules low profit jobs first.

This leads to suboptimal total profit and incorrect scheduling.

Sorting should be descending by profit.

📝 Syntax

advanced

Identify Syntax Error in Job Scheduling Code

Which option contains a syntax error in the TypeScript code for job scheduling?

DSA Typescript

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 }
];

function scheduleJobs(jobs: Job[]): string {
  jobs.sort((a, b) => b.profit - a.profit);
  const n = jobs.length;
  const result: (string | null)[] = Array(n).fill(null);
  const slot: boolean[] = Array(n).fill(false);

  for (let i = 0; i < n; i++) {
    for (let j = Math.min(n - 1, jobs[i].deadline - 1); j >= 0; j--) {
      if (!slot[j]) {
        slot[j] = true;
        result[j] = jobs[i].id;
        break;
      }
    }
  }

  return result.filter(x => x !== null).join(' -> ') + ' -> null';
}

A Missing semicolon after slot[j] = true inside the inner loop.

B Incorrect type annotation for result array; should be string[] not (string | null)[]

C Using Math.min with n - 1 and jobs[i].deadline - 1 causes runtime error.

D The function scheduleJobs is missing a return type annotation.

💡 Hint

Look carefully at statement endings inside loops.

TypeScript allows optional semicolons but missing semicolon after slot[j] = true can cause issues in some cases.

Other options are valid TypeScript syntax.

🚀 Application

expert

Optimal Job Scheduling Output for Complex Input

Given the following jobs with deadlines and profits, what is the optimal job schedule output string? Jobs: - {id: 'j1', deadline: 4, profit: 70} - {id: 'j2', deadline: 1, profit: 80} - {id: 'j3', deadline: 1, profit: 30} - {id: 'j4', deadline: 2, profit: 100} - {id: 'j5', deadline: 3, profit: 40} - {id: 'j6', deadline: 2, profit: 20} Use the greedy algorithm to schedule jobs to maximize profit. Each job takes 1 unit time. Output format: job ids separated by ' -> ' ending with ' -> null'.

A "j2 -> j4 -> j5 -> j1 -> null"

B "j4 -> j2 -> j5 -> j1 -> null"

C "j4 -> j5 -> j1 -> j2 -> null"

D "j2 -> j4 -> j1 -> j5 -> null"

💡 Hint

Sort jobs by profit descending and assign each to the latest free slot before deadline.

Sorted by profit: j4(100), j2(80), j1(70), j5(40), j3(30), j6(20).

Assign j4 to slot 1 (deadline 2), j2 to slot 0 (deadline 1), j5 to slot 2 (deadline 3), j1 to slot 3 (deadline 4).

Slots 0 to 3 correspond to time 1 to 4.

Schedule is "j4 -> j2 -> j5 -> j1 -> null".

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