Job Scheduling with Deadlines in DSA C - 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 changes as the number of jobs grows.

How does the scheduling process scale when we add more jobs with deadlines?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Job Scheduling with Deadlines
// jobs[] contains pairs of (deadline, profit)
// n is number of jobs
void jobScheduling(Job jobs[], int n) {
    // Sort jobs by profit descending
    qsort(jobs, n, sizeof(Job), compareProfit);

    int result[n];
    bool slot[n];
    for (int i = 0; i < n; i++) slot[i] = false;

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

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops where for each job we try to find a free slot before its deadline.
• How many times: Outer loop runs n times (for each job), inner loop can run up to n times in worst case (checking slots).

How Execution Grows With Input

As the number of jobs increases, the number of checks to find free slots grows roughly with the square of the number of jobs.

Input Size (n)	Approx. Operations
10	About 100 checks
100	About 10,000 checks
1000	About 1,000,000 checks

Pattern observation: The operations grow roughly by the square of the input size.

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 scheduling always takes linear time because we just loop through jobs once."

[OK] Correct: Each job may require checking multiple slots before finding a free one, causing nested loops and more work as jobs increase.

Interview Connect

Understanding how nested loops affect time helps you explain and improve scheduling algorithms clearly in real situations.

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 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