Job Scheduling with Deadlines in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a vector of jobs with their deadlines and profits.

DSA Rust

let jobs = vec![("Job1", 2, 100), ("Job2", [1], 19), ("Job3", 2, 27)];

Drag options to blanks, or click blank then click option'

A3

B1

C4

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using a deadline less than or equal to the first job's deadline.

Confusing profit with deadline.

The second job has a deadline of 3, so the correct value is 3.

2fill in blank

medium

Complete the code to sort jobs by profit in descending order.

DSA Rust

jobs.sort_by(|a, b| b.[1].cmp(&a.[1]));

Drag options to blanks, or click blank then click option'

A2

B1

C0

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong index for profit.

Sorting in ascending order instead of descending.

The profit is the third element (index 2) in the tuple, so we compare b.2 with a.2.

3fill in blank

hard

Fix the error in the code to initialize the result vector with None values for scheduling.

DSA Rust

let mut result: Vec<Option<&str>> = vec![[1]; max_deadline];

Drag options to blanks, or click blank then click option'

A""

BNone

CSome("")

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or empty string instead of None.

Using Some("") which is not empty.

We need to initialize the vector with None values to represent empty slots.

4fill in blank

hard

Fill both blanks to check if a slot is free and assign the job to that slot.

DSA Rust

if result[[1]].is_none() {
    result[[2]] = Some(job.0);
}

Drag options to blanks, or click blank then click option'

Aslot

Bi

Cj

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for checking and assigning.

Using undefined variables.

We use the same variable 'i' to check and assign the slot in the result vector.

5fill in blank

hard

Fill all three blanks to complete the loop that schedules jobs before their deadlines.

DSA Rust

for job in jobs.iter() {
    let mut [1] = job.1.min(max_deadline) - 1;
    while [2] >= 0 {
        if result[[3]].is_none() {
            result[[3]] = Some(job.0);
            break;
        }
        [2] -= 1;
    }
}

Drag options to blanks, or click blank then click option'

Aslot

Bi

Cindex

Dpos

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for the same purpose.

Not decrementing the index correctly.

The variable 'i' is used to track the slot index for scheduling jobs.

Practice

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

easy

A. Schedule jobs in the order they arrive without considering deadlines

B. Minimize the number of jobs scheduled

C. Maximize total profit by scheduling jobs before their deadlines

D. Maximize the number of jobs scheduled regardless of profit

Solution

1. Step 1: Understand the problem goal

   The problem asks to schedule jobs so that each job finishes before its deadline and the total profit is maximized.

2. Step 2: Identify the objective

   Maximizing total profit means choosing jobs with higher profit and fitting them before their deadlines.

3. Final Answer:

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

4. Quick Check:

   Goal = Maximize profit before deadlines [OK]

Hint: Focus on profit maximization within deadlines [OK]

Common Mistakes:

• Thinking the goal is to schedule all jobs
• Ignoring deadlines when scheduling
• Trying to minimize jobs instead of maximizing profit

2. Which Rust data structure is best suited to keep track of free time slots when implementing job scheduling with deadlines?

easy

A. Vec (vector) with boolean flags for slots

B. HashMap with job IDs as keys

C. LinkedList of jobs sorted by profit

D. BinaryHeap for storing jobs by deadline

Solution

1. Step 1: Identify the need for tracking free slots

   We need to know which time slots are free to schedule jobs before their deadlines.

2. Step 2: Choose a data structure for slot tracking

   A vector of booleans can mark each slot as free or occupied efficiently.

3. Final Answer:

   Vec (vector) with boolean flags for slots -> Option A

4. Quick Check:

   Use Vec<bool> to track free slots [OK]

Hint: Use a boolean vector to mark free/occupied slots [OK]

Common Mistakes:

• Using HashMap which is unnecessary for fixed slots
• Using LinkedList which is inefficient for slot checking
• Using BinaryHeap which is for priority, not slot tracking

3. What is the output of the following Rust code snippet that schedules jobs with deadlines and profits?

struct Job { id: char, deadline: usize, profit: usize }

fn job_scheduling(jobs: &mut Vec<Job>) -> Vec<char> {
    jobs.sort_by(|a, b| b.profit.cmp(&a.profit));
    let max_deadline = jobs.iter().map(|j| j.deadline).max().unwrap_or(0);
    let mut slots = vec![false; max_deadline];
    let mut result = Vec::new();

    for job in jobs.iter() {
        for slot in (0..job.deadline).rev() {
            if !slots[slot] {
                slots[slot] = true;
                result.push(job.id);
                break;
            }
        }
    }
    result
}

fn main() {
    let mut jobs = vec![
        Job { id: 'A', deadline: 2, profit: 100 },
        Job { id: 'B', deadline: 1, profit: 19 },
        Job { id: 'C', deadline: 2, profit: 27 },
        Job { id: 'D', deadline: 1, profit: 25 },
        Job { id: 'E', deadline: 3, profit: 15 },
    ];
    let scheduled = job_scheduling(&mut jobs);
    println!("{:?}", scheduled);
}

medium

A. ['A', 'B', 'E']

B. ['A', 'D', 'E']

C. ['B', 'D', 'E']

D. ['A', 'C', 'E']

Solution

1. Step 1: Sort jobs by profit descending

   Sorted jobs: A(100,2), C(27,2), D(25,1), B(19,1), E(15,3)

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

   Schedule A at slot 1 (0-based), C at slot 0, E at slot 2. D and B cannot be scheduled as slots 0 and 1 are taken.

3. Final Answer:

   ['A', 'C', 'E'] -> Option D

4. Quick Check:

   Jobs scheduled = ['A', 'C', 'E'] [OK]

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

Common Mistakes:

• Scheduling jobs in arrival order instead of profit order
• Assigning earliest slot instead of latest free slot
• Ignoring slot availability and deadlines

4. Identify the bug in this Rust code snippet for job scheduling with deadlines:

fn schedule_jobs(jobs: &mut Vec<(char, usize, usize)>) -> Vec<char> {
    jobs.sort_by(|a, b| b.2.cmp(&a.2));
    let max_deadline = jobs.iter().map(|j| j.1).max().unwrap_or(0);
    let mut slots = vec![false; max_deadline];
    let mut result = Vec::new();

    for job in jobs.iter() {
        for slot in 0..job.1 {
            if !slots[slot] {
                slots[slot] = true;
                result.push(job.0);
                break;
            }
        }
    }
    result
}

medium

A. Jobs are sorted by profit ascending instead of descending

B. Inner loop should iterate slots in reverse order to assign latest slot

C. Result vector should store job profits instead of IDs

D. Slots vector size is incorrect, should be max_deadline + 1

Solution

1. Step 1: Check sorting order

   Jobs are sorted by profit descending (b.2.cmp(&a.2)), which is correct to prioritize high-profit jobs.

2. Step 2: Check slot assignment loop

   The loop iterates from 0 to deadline-1, assigning the earliest available slot. The correct approach assigns the latest free slot before the deadline to leave earlier slots for future jobs with tighter deadlines, so the loop should be reversed using (0..job.1).rev().

3. Final Answer:

   Inner loop should iterate slots in reverse order to assign latest slot -> Option B

4. Quick Check:

   Assign latest free slot by looping backwards [OK]

Hint: Loop slots backward to assign latest free slot [OK]

Common Mistakes:

• Assigning earliest slot instead of latest free slot
• Using wrong slot vector size (off-by-one)
• Storing wrong data in result vector

5. Given jobs with deadlines and profits: A(4, 70), B(1, 80), C(1, 30), D(2, 100), E(3, 40), how many jobs can be scheduled to maximize profit and what is the total profit? Use the job scheduling with deadlines approach.

hard

A. 4 jobs, total profit 290

B. 3 jobs, total profit 250

C. 3 jobs, total profit 220

D. 4 jobs, total profit 240

Solution

1. Step 1: Sort jobs by profit descending

   Sorted: D(100,2), B(80,1), A(70,4), E(40,3), C(30,1)

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

   Schedule B at slot 0 (deadline 1), D at slot 1 (deadline 2), E at slot 2 (deadline 3), A at slot 3 (deadline 4). C cannot be scheduled as slot 0 is taken.

3. Step 3: Calculate total profit

   Profits: B(80) + D(100) + E(40) + A(70) = 290

4. Final Answer:

   4 jobs, total profit 290 -> Option A

5. Quick Check:

   4 jobs scheduled, profit = 290 [OK]

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

Common Mistakes:

• Scheduling fewer jobs than possible
• Ignoring slot availability and deadlines
• Adding profits of unscheduled jobs