What if you could always finish the most important tasks before their deadlines without stress?
Why Job Scheduling with Deadlines in DSA Typescript?
Imagine you have many tasks to do today, each with a time limit to finish. You try to pick which task to do first by guessing, writing notes on paper, and constantly checking if you can finish on time.
This guessing and checking wastes time and causes mistakes. You might miss important tasks or run out of time because you didn't plan well. It feels like juggling too many things without a clear order.
Job Scheduling with Deadlines helps you plan tasks smartly. It picks the best tasks to do first so you finish as many as possible before their deadlines. This way, you never waste time guessing and always get the most done.
const tasks = [{id:1, deadline:4, profit:10}, {id:2, deadline:1, profit:20}];
// Manually pick tasks and check deadlines one by one
for (let task of tasks) {
if (canFinish(task)) doTask(task);
}const tasks = [{id:1, deadline:4, profit:10}, {id:2, deadline:1, profit:20}];
tasks.sort((a,b) => b.profit - a.profit);
for (let task of tasks) {
if (slotAvailable(task.deadline)) schedule(task);
}You can efficiently schedule many tasks to maximize your success without stress or guesswork.
Delivery companies use job scheduling to assign packages so drivers deliver the most packages before promised times.
Manual task planning is slow and error-prone.
Job Scheduling with Deadlines finds the best order to do tasks.
This method helps finish maximum tasks on time easily.
Practice
Job Scheduling with Deadlines problem?Solution
Step 1: Understand the problem goal
The problem asks to schedule jobs so that total profit is maximized while respecting deadlines.Step 2: Identify the objective
Scheduling jobs before their deadlines to get maximum profit is the main goal.Final Answer:
Maximize the total profit by scheduling jobs before their deadlines -> Option DQuick Check:
Goal = Maximize profit with deadlines [OK]
- Ignoring deadlines and scheduling all jobs
- Trying to minimize jobs instead of maximizing profit
- Scheduling jobs in input order without sorting
id, deadline, and profit properties?Solution
Step 1: Check correct TypeScript interface syntax
interface Job { id: number; deadline: number; profit: number; } correctly declares an interface with typed properties.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.Final Answer:
interface Job { id: number; deadline: number; profit: number; } -> Option CQuick Check:
Correct interface syntax = interface Job { id: number; deadline: number; profit: number; } [OK]
- Using class without types for data structure
- Omitting types in interface properties
- Using wrong types like string for numeric fields
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));Solution
Step 1: Sort jobs by profit descending
Sorted jobs: a(100), c(27), d(25), b(19), e(15).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).Final Answer:
['a', 'c', 'e'] -> Option BQuick Check:
Scheduled jobs = ['a', 'c', 'e'] [OK]
- Scheduling jobs in input order without sorting
- Assigning earliest slot instead of latest
- Ignoring deadline constraints
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[];
}Solution
Step 1: Analyze sorting order
Jobs are sorted by profit ascending, which schedules low profit jobs first, reducing total profit.Step 2: Correct sorting order
Sorting by profit descending ensures high profit jobs are scheduled first for maximum profit.Final Answer:
Sort jobs by profit descending instead of ascending -> Option AQuick Check:
Profit descending sort fixes scheduling [OK]
- Sorting by ascending profit reduces total profit
- Changing loop direction incorrectly
- Using zero instead of null for empty slots
Solution
Step 1: Sort jobs by profit descending
Order: Job4(100), Job2(80), Job1(70), Job5(40), Job3(30).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).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.Step 4: Re-check options
Confirm scheduling: slots 0:Job2, 1:Job4, 2:Job5, 3:Job1. All fit, total 290.Final Answer:
290 -> Option AQuick Check:
Max profit = 290 [OK]
- Assuming all jobs fit without checking slots
- Adding profits of unscheduled jobs
- Misinterpreting deadlines as exact time slots
Ready to prove it?
Pass the test and this pattern moves unknown
