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 we add more jobs.

Specifically, how does the scheduling process grow when the number of jobs increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <vector>
#include <algorithm>
using namespace std;

struct Job {
  int id, deadline, profit;
};

void scheduleJobs(vector<Job>& jobs) {
  sort(jobs.begin(), jobs.end(), [](Job a, Job b) { return a.profit > b.profit; });
  vector<int> result(jobs.size(), -1);
  vector<bool> slot(jobs.size(), false);
  for (int i = 0; i < jobs.size(); i++) {
    for (int j = min((int)jobs.size() - 1, 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 available slot before its deadline.

Identify Repeating Operations

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

How Execution Grows With Input

As the number of jobs increases, the sorting step takes more time, and the nested loops try more slots.

Input Size (n)	Approx. Operations
10	About 100 operations (sorting + nested loops)
100	About 10,000 operations
1000	About 1,000,000 operations

Pattern observation: The operations grow roughly with the square of the number of jobs.

Final Time Complexity

Time Complexity: O(n^2)

This means if you double the number of jobs, the time needed roughly quadruples.

Common Mistake

[X] Wrong: "Since we only loop through jobs once, the time is O(n)."

[OK] Correct: Each job may check multiple slots in the inner loop, causing nested loops and more work than just one pass.

Interview Connect

Understanding how nested loops affect time helps you explain your code clearly and shows you can analyze efficiency in real tasks.

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. Minimize the number of jobs scheduled

B. Schedule jobs in the order they appear without considering deadlines

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 objective

   The problem aims to schedule jobs so that each job finishes before its deadline to earn profit.

2. Step 2: Identify the optimization goal

   The goal is to maximize the total profit earned by selecting jobs that fit within their deadlines.

3. Final Answer:

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

4. Quick Check:

   Goal = Maximize profit [OK]

Hint: Focus on profit maximization within deadlines [OK]

Common Mistakes:

• Ignoring deadlines and scheduling all jobs
• Trying to schedule jobs in input order
• Maximizing number of jobs instead of profit

2. Which of the following is the correct C++ syntax to sort jobs by profit in descending order using std::sort and a custom comparator?

easy

A. std::sort(jobs.begin(), jobs.end());

B. std::sort(jobs.begin(), jobs.end(), [](Job a, Job b){ return a.profit > b.profit; });

C. std::sort(jobs.begin(), jobs.end(), compare); // where compare returns a.profit < b.profit

D. std::sort(jobs.begin(), jobs.end(), [](Job a, Job b){ return a.profit < b.profit; });

Solution

1. Step 1: Understand sorting by profit descending

   To sort descending, comparator should return true if first profit is greater than second.

2. Step 2: Check lambda comparator syntax

   std::sort(jobs.begin(), jobs.end(), [](Job a, Job b){ return a.profit > b.profit; }); uses a lambda returning a.profit > b.profit, which sorts descending correctly.

3. Final Answer:

   std::sort(jobs.begin(), jobs.end(), [](Job a, Job b){ return a.profit > b.profit; }); -> Option B

4. Quick Check:

   Descending sort = a.profit > b.profit [OK]

Hint: Use comparator returning true if first > second for descending [OK]

Common Mistakes:

• Using < instead of > for descending order
• Not providing a comparator for custom sorting
• Using default sort which sorts ascending

3. Given jobs with deadlines and profits:
jobs = [{id:1, deadline:2, profit:100}, {id:2, deadline:1, profit:19}, {id:3, deadline:2, profit:27}, {id:4, deadline:1, profit:25}, {id:5, deadline:3, profit:15}]
What is the maximum profit achievable by scheduling these jobs?

medium

A. 185

B. 190

C. 160

D. 142

Solution

1. Step 1: Sort jobs by profit descending

   Sorted jobs: (1,100,2), (3,27,2), (4,25,1), (2,19,1), (5,15,3)

2. Step 2: Schedule jobs greedily before deadlines

   Schedule job 1 at slot 2, job 4 at slot 1, job 5 at slot 3. Jobs 3 and 2 cannot fit without conflict.

3. Step 3: Calculate total profit

   Profit = 100 + 25 + 15 = 140. But check if job 3 can replace job 4 for better profit: job 3 (27) > job 4 (25), so schedule job 3 at slot 2, job 4 at slot 1 is not possible because slot 1 is taken by job 4. Instead, schedule job 3 at slot 2, job 2 at slot 1 (profit 19), job 5 at slot 3 (15). Total = 100 + 27 + 15 = 142. Better is job 1 (100), job 3 (27), job 4 (25) at slots 2,1,3 respectively? Slot 3 deadline is 3, so job 4 deadline 1 can't be at 3. So best is job 1 (100), job 3 (27), job 4 (25) at slots 2,1, no slot for job 5. Total 152. Actually, slot 3 is free for job 5 (15). So schedule job 1 at 2, job 4 at 1, job 5 at 3 = 100+25+15=140. Or job 1 at 2, job 3 at 1, job 5 at 3 = 100+27+15=142. Or job 1 at 2, job 4 at 1, job 3 can't fit. So max profit is 142.

4. Final Answer:

   142 -> Option D

5. Quick Check:

   Max profit = 142 [OK]

Hint: Sort by profit, schedule greedily before deadlines [OK]

Common Mistakes:

• Scheduling jobs ignoring deadlines
• Not sorting by profit first
• Overlapping jobs in same time slot

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

struct Job { int id, deadline, profit; };
std::vector<Job> jobs = ...;
std::sort(jobs.begin(), jobs.end(), [](Job a, Job b){ return a.profit < b.profit; });
std::vector<int> slot(maxDeadline + 1, -1);
for (auto &job : jobs) {
  for (int j = job.deadline; j > 0; j--) {
    if (slot[j] == -1) {
      slot[j] = job.id;
      break;
    }
  }
}

medium

A. Sorting comparator sorts ascending instead of descending

B. Slots vector size should be maxDeadline, not maxDeadline + 1

C. Inner loop should run from 1 to job.deadline, not reverse

D. Job IDs should not be stored in slots, profits should be stored

Solution

1. Step 1: Analyze sorting comparator

   The comparator returns true if a.profit < b.profit, which sorts ascending by profit.

2. Step 2: Understand scheduling logic

   Jobs must be sorted descending by profit to schedule highest profit jobs first.

3. Final Answer:

   Sorting comparator sorts ascending instead of descending -> Option A

4. Quick Check:

   Comparator must sort descending [OK]

Hint: Sort jobs descending by profit for correct scheduling [OK]

Common Mistakes:

• Using ascending sort comparator
• Incorrect slot vector size off by one
• Loop direction confusion

5. You have 6 jobs with deadlines and profits:
jobs = [{id:1, d:4, p:70}, {id:2, d:1, p:80}, {id:3, d:1, p:30}, {id:4, d:1, p:100}, {id:5, d:3, p:40}, {id:6, d:2, p:20}]
Using the job scheduling algorithm, what is the maximum profit and which jobs are scheduled?

hard

A. Profit: 230; Jobs scheduled: 4 -> 1 -> 5 -> 6

B. Profit: 220; Jobs scheduled: 4 -> 2 -> 5 -> 6

C. Profit: 240; Jobs scheduled: 4 -> 2 -> 1 -> 5

D. Profit: 210; Jobs scheduled: 4 -> 1 -> 3 -> 5

Solution

1. Step 1: Sort jobs by profit descending

   Sorted: (4,100,1), (2,80,1), (1,70,4), (5,40,3), (3,30,1), (6,20,2)

2. Step 2: Schedule jobs greedily before deadlines

   Schedule job 4 at slot 1 (deadline 1), job 6 at slot 2 (deadline 2), job 5 at slot 3 (deadline 3), job 1 at slot 4 (deadline 4). Jobs 2 and 3 cannot fit because slot 1 is taken by job 4.

3. Step 3: Calculate total profit and jobs scheduled

   Profit = 100 + 20 + 40 + 70 = 230; Scheduled jobs in order of slots: 4 -> 6 -> 5 -> 1 (slots 1,2,3,4)

4. Final Answer:

   Profit: 230; Jobs scheduled: 4 -> 6 -> 5 -> 1 -> Option A

5. Quick Check:

   Max profit = 230 with jobs 4,6,5,1 [OK]

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

Common Mistakes:

• Scheduling multiple jobs in same slot
• Ignoring deadlines when scheduling
• Not sorting jobs by profit first