Job Scheduling with Deadlines in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the Job structure with id and deadline fields.

DSA C

typedef struct Job {
    int id;
    int [1];
} Job;

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

Apriority

Bduration

Cdeadline

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'priority' instead of 'deadline' for the deadline field.

Confusing 'duration' with deadline.

The field to store the deadline of a job is named deadline.

2fill in blank

medium

Complete the code to sort jobs by their deadline in ascending order.

DSA C

int compareJobs(const void *a, const void *b) {
    Job *jobA = (Job *)a;
    Job *jobB = (Job *)b;
    return jobA->[1] - jobB->[1];
}

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

Aduration

Bdeadline

Cpriority

Did

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by id instead of deadline.

Using priority or duration which are not relevant here.

Sorting by deadline ensures jobs with earlier deadlines come first.

3fill in blank

hard

Fix the error in the code to schedule jobs within their deadlines using a greedy approach.

DSA C

int scheduleJobs(Job jobs[], int result[], int n) {
    int slot[n];
    for (int i = 0; i < n; i++)
        slot[i] = 0;

    for (int i = 0; i < n; i++) {
        for (int j = jobs[i].deadline - 1; j >= 0; j--) {
            if (slot[j] == [1]) {
                result[j] = i;
                slot[j] = 1;
                break;
            }
        }
    }
    return 0;
}

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A0

B1

C-1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if slot[j] == 1 which means occupied, so no assignment happens.

Using negative or invalid values for slot check.

The slot array uses 0 to indicate a free slot and 1 for occupied. We check if slot[j] == 0 to find a free slot.

4fill in blank

hard

Fill both blanks to print the scheduled jobs in order.

DSA C

void printSchedule(Job jobs[], int result[], int n) {
    printf("Scheduled jobs: ");
    for (int i = 0; i < n; i++) {
        if (result[i] != [1]) {
            printf("%d ", jobs[result[i]].[2]);
        }
    }
    printf("\n");
}

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A-1

B0

Cid

Ddeadline

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for 0 instead of -1 in result array.

Printing deadline instead of job id.

We check if result[i] is not -1 (meaning a job is scheduled) and print the job's id.

5fill in blank

hard

Fill all three blanks to initialize the result array with -1 and sort jobs by deadline.

DSA C

void initializeAndSort(Job jobs[], int n, int result[]) {
    for (int i = 0; i < n; i++) {
        result[i] = [1];
    }
    qsort(jobs, n, sizeof(Job), [2]);
}

int compareByDeadline(const void *a, const void *b) {
    Job *jobA = (Job *)a;
    Job *jobB = (Job *)b;
    return jobA->[3] - jobB->[3];
}

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

A-1

BcompareByDeadline

Cdeadline

Did

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing result with 0 instead of -1.

Using wrong compare function name in qsort.

Comparing by id instead of deadline.

We initialize result with -1 to mark empty slots, use compareByDeadline for sorting, and compare the deadline field.

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 within their deadlines

Solution

1. Step 1: Understand the problem objective

   The problem asks to schedule jobs so that the total profit is maximized while respecting each job's deadline.

2. Step 2: Identify the goal

   The goal is to select jobs that fit within their deadlines and yield the highest total profit.

3. Final Answer:

   Maximize the total profit by scheduling jobs within their deadlines -> Option D

4. Quick Check:

   Goal = Maximize profit within deadlines [OK]

Hint: Focus on maximizing profit while meeting deadlines [OK]

Common Mistakes:

• Ignoring deadlines and scheduling all jobs
• Trying to minimize jobs instead of maximizing profit
• Scheduling jobs without considering profit

2. Which of the following is the correct way to sort jobs by profit in descending order in C?

easy

A. sort(jobs, n, sizeof(Job), compareProfitDesc);

B. sort(jobs, jobs+n, compareProfitDesc);

C. qsort(jobs, n, sizeof(Job), compareProfitDesc);

D. qsort(jobs, n, sizeof(Job), compareProfitAsc);

Solution

1. Step 1: Identify the sorting function in C

   In C, qsort is used for sorting arrays with a comparator function.

2. Step 2: Check the correct parameters

   The correct call is qsort(array, number_of_elements, size_of_element, comparator_function). qsort(jobs, n, sizeof(Job), compareProfitDesc); matches this.

3. Final Answer:

   qsort(jobs, n, sizeof(Job), compareProfitDesc); -> Option C

4. Quick Check:

   Use qsort with correct parameters in C [OK]

Hint: Use qsort with array, count, size, comparator in C [OK]

Common Mistakes:

• Using C++ sort() in C code
• Passing wrong size or count parameters
• Using ascending comparator instead of descending

3. Given jobs with profits and deadlines: {Job1: profit=20, deadline=2}, {Job2: profit=15, deadline=2}, {Job3: profit=10, deadline=1}, {Job4: profit=5, deadline=3}, what is the maximum total profit after scheduling?

medium

A. 40

B. 45

C. 50

D. 35

Solution

1. Step 1: Sort jobs by profit descending

   Order: Job1(20,2), Job2(15,2), Job3(10,1), Job4(5,3)

2. Step 2: Schedule jobs greedily within deadlines

   Schedule Job1 at slot 2, Job2 at slot 1 (since slot 2 taken), Job4 at slot 3. Job3 cannot be scheduled (slot 1 taken). Total profit = 20+15+5=40.

3. Final Answer:

   40 -> Option A

4. Quick Check:

   Greedy scheduling = 40 profit [OK]

Hint: Schedule highest profit jobs first within deadlines [OK]

Common Mistakes:

• Scheduling jobs ignoring deadlines
• Overlapping jobs in same slot
• Counting all jobs without checking slot availability

4. Identify the error in this C code snippet for job scheduling:

for (int i = 0; i < n; i++) {
  int slot = jobs[i].deadline - 1;
  while (slot >= 0 && result[slot] != -1) {
    slot--;
  }
  if (slot >= 0) {
    result[slot] = i;
  }
}

medium

A. The loop should start from the last job, not the first

B. The array 'result' is not initialized to -1 before use

C. The condition 'slot >= 0' should be 'slot > 0'

D. The deadline should not be decremented by 1

Solution

1. Step 1: Check initialization of result array

   The array 'result' stores scheduled jobs. It must be initialized to -1 to mark free slots.

2. Step 2: Identify missing initialization

   The code snippet does not show initialization of 'result' to -1, which can cause incorrect scheduling.

3. Final Answer:

   The array 'result' is not initialized to -1 before use -> Option B

4. Quick Check:

   Initialize result array to -1 before scheduling [OK]

Hint: Always initialize scheduling array slots to -1 [OK]

Common Mistakes:

• Forgetting to initialize result array
• Wrong loop direction for scheduling
• Incorrect slot boundary conditions

5. You have 4 jobs: J1(profit=100, deadline=2), J2(profit=19, deadline=1), J3(profit=27, deadline=2), J4(profit=25, deadline=1). Using the job scheduling algorithm, what is the maximum profit and which jobs are scheduled?

hard

A. Profit=127; Jobs scheduled: J1 at slot 2, J3 at slot 1

B. Profit=144; Jobs scheduled: J1 at slot 2, J3 at slot 1, J2 at slot 1

C. Profit=125; Jobs scheduled: J1 at slot 1, J4 at slot 2

D. Profit=100; Jobs scheduled: J1 only

Solution

1. Step 1: Sort jobs by profit descending

   Order: J1(100,2), J3(27,2), J4(25,1), J2(19,1)

2. Step 2: Schedule jobs greedily within deadlines

   Schedule J1 at slot 2, J3 at slot 1 (slot 1 free). J4 and J2 cannot be scheduled as slot 1 is taken. Total profit = 100 + 27 = 127.

3. Final Answer:

   Profit=127; Jobs scheduled: J1 at slot 2, J3 at slot 1 -> Option A

4. Quick Check:

   Greedy scheduling = 127 profit [OK]

Hint: Schedule highest profit jobs first, fill latest slots [OK]

Common Mistakes:

• Scheduling multiple jobs in same slot
• Ignoring deadlines when scheduling
• Adding profits of unscheduled jobs