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

Time Complexity: Job Scheduling with Deadlines

O(n * d)

Understanding Time Complexity

We want to understand how the time needed to schedule jobs grows as we add more jobs.

How does the number of jobs affect the work done by the scheduling algorithm?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn job_scheduling(jobs: &mut Vec<(i32, i32)>) -> i32 {
    jobs.sort_by(|a, b| b.0.cmp(&a.0));
    let mut result = vec![-1; 100];
    let mut count = 0;
    for &(profit, deadline) in jobs.iter() {
        for slot in (0..deadline).rev() {
            if result[slot as usize] == -1 {
                result[slot as usize] = profit;
                count += profit;
                break;
            }
        }
    }
    count
}

This code schedules jobs by profit and deadline to maximize total profit.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops - outer loop over all jobs, inner loop over possible time slots.
• How many times: Outer loop runs once per job (n times), inner loop runs up to the deadline (up to 100 times).

How Execution Grows With Input

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

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

Pattern observation: The work grows roughly by the product of the number of jobs and the maximum deadline.

Final Time Complexity

Time Complexity: O(n * d) where n is number of jobs and d is maximum deadline (here fixed at 100).

This means if you double the number of jobs, the time needed roughly doubles, assuming deadline is fixed.

Common Mistake

[X] Wrong: "The algorithm only loops once over the jobs, so it must be O(n)."

[OK] Correct: Because inside the job loop, it also loops over time slots, which can be up to the maximum deadline times, making it slower than just one loop.

Interview Connect

Understanding how nested loops affect time helps you explain your code clearly and shows you can think about 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. 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