Job Scheduling with Deadlines in DSA Typescript - Time & Space Complexity

Time Complexity: Job Scheduling with Deadlines

O(n^2)

Understanding Time Complexity

We want to understand how the time needed to schedule jobs grows as the number of jobs increases.

How does the code handle more jobs and deadlines efficiently?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function jobScheduling(jobs: {id: number, deadline: number, profit: number}[]): number[] {
  jobs.sort((a, b) => b.profit - a.profit);
  const result: number[] = [];
  const slot: boolean[] = Array(jobs.length).fill(false);
  for (const job of jobs) {
    for (let j = Math.min(jobs.length, job.deadline) - 1; j >= 0; j--) {
      if (!slot[j]) {
        slot[j] = true;
        result.push(job.id);
        break;
      }
    }
  }
  return result;
}
    

This code schedules jobs to maximize profit by placing each job in the latest available slot before its deadline.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops where the outer loop goes through each job and the inner loop searches for a free slot before the job's deadline.
• How many times: Outer loop runs once per job (n times), inner loop can run up to n times in worst case.

How Execution Grows With Input

As the number of jobs grows, the code checks more jobs and searches more slots for each job.

Input Size (n)	Approx. Operations
10	Up to 100 checks (10 jobs x 10 slots)
100	Up to 10,000 checks (100 jobs x 100 slots)
1000	Up to 1,000,000 checks (1000 jobs x 1000 slots)

Pattern observation: The number of operations grows roughly with the square of the number of jobs.

Final Time Complexity

Time Complexity: O(n2)

This means if you double the number of jobs, the time to schedule them roughly quadruples.

Common Mistake

[X] Wrong: "The inner loop only runs once per job, so the code is O(n)."

[OK] Correct: Sometimes the inner loop must check many slots before finding a free one, especially when deadlines overlap, causing many repeated checks.

Interview Connect

Understanding how nested loops affect time helps you explain your approach clearly and shows you can analyze efficiency in real problems.

Self-Check

"What if we used a data structure to find free slots faster? How would the time complexity change?"

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