Job Scheduling with Deadlines in DSA Go - 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 grows as we add more jobs.

Specifically, how the steps increase when we try to fit jobs before their deadlines.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

package main

import "sort"

type Job struct {
    id       int
    deadline int
    profit   int
}

func JobScheduling(jobs []Job) []int {
    sort.Slice(jobs, func(i, j int) bool { return jobs[i].profit > jobs[j].profit })
    n := len(jobs)
    result := make([]int, n)
    slot := make([]bool, n)

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

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

This code schedules jobs by profit, 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 the outer loop goes through each job and the inner loop searches for a free slot before the job's deadline.
• How many times: Outer loop runs n times (number of jobs), inner loop can run up to n times in worst case.

How Execution Grows With Input

As the number of jobs grows, the code tries to place each job by checking slots backward from its deadline.

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

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

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 searching backward through slots, which can take up to n steps, making the total work grow faster than just one loop.

Interview Connect

Understanding how nested loops affect time helps you explain your approach clearly and shows you can reason 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 appear without considering deadlines

B. Minimize the number of jobs scheduled

C. Maximize the 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 and the total profit is maximized.

2. Step 2: Identify the correct goal

   Maximizing total profit while respecting deadlines is the key goal, not just scheduling all jobs or ignoring deadlines.

3. Final Answer:

   Maximize the 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:

• Ignoring deadlines and scheduling all jobs
• Trying to minimize jobs instead of maximizing profit
• Scheduling jobs in input order without sorting

2. Which Go data structure is best suited to efficiently pick the next available time slot when scheduling jobs by deadline?

easy

A. Map (hash table)

B. Disjoint Set (Union-Find)

C. Slice (dynamic array)

D. Linked List

Solution

1. Step 1: Understand the scheduling requirement

   We need to quickly find and assign the latest free slot before a job's deadline.

2. Step 2: Identify the suitable data structure

   Disjoint Set (Union-Find) efficiently tracks and merges available slots, making it ideal for this task.

3. Final Answer:

   Disjoint Set (Union-Find) -> Option B

4. Quick Check:

   Best structure for slot tracking = Disjoint Set [OK]

Hint: Use Disjoint Set to find free slots fast [OK]

Common Mistakes:

• Using slices which require linear search
• Using maps which don't track slot unions
• Using linked lists which are inefficient for this

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

package main
import "fmt"
type Job struct {
  id     string
  deadline int
  profit int
}
func main() {
  jobs := []Job{{"a", 2, 100}, {"b", 1, 19}, {"c", 2, 27}, {"d", 1, 25}, {"e", 3, 15}}
  // Assume jobs are sorted by profit descending
  scheduled := []string{"a", "c", "e"}
  fmt.Println(scheduled)
}

medium

A. ["a", "c", "e"]

B. ["b", "d", "e"]

C. ["a", "b", "c"]

D. ["a", "d", "e"]

Solution

1. Step 1: Analyze the job list and sorting

   Jobs are sorted by profit descending: a(100), c(27), d(25), b(19), e(15). The code assumes scheduled jobs are a, c, e.

2. Step 2: Check feasibility of scheduled jobs

   Job a (deadline 2), c (deadline 2), e (deadline 3) can be scheduled without conflict, maximizing profit.

3. Final Answer:

   ["a", "c", "e"] -> Option A

4. Quick Check:

   Scheduled jobs = ["a", "c", "e"] [OK]

Hint: Pick jobs by profit, fit before deadlines [OK]

Common Mistakes:

• Ignoring deadlines and picking all high profit jobs
• Confusing job order in output
• Assuming jobs b or d are scheduled incorrectly

4. Identify the bug in this Go code snippet for job scheduling:

func scheduleJobs(jobs []Job) []string {
  sort.Slice(jobs, func(i, j int) bool {
    return jobs[i].profit > jobs[j].profit
  })
  result := []string{}
  slots := make([]bool, len(jobs))
  for _, job := range jobs {
    for j := job.deadline; j >= 0; j-- {
      if !slots[j] {
        slots[j] = true
        result = append(result, job.id)
        break
      }
    }
  }
  return result
}

medium

A. Slots array size should be job.deadline, not len(jobs)

B. Jobs should be sorted by deadline, not profit

C. Loop index j should not go below 1

D. Loop index j should start from job.deadline - 1

Solution

1. Step 1: Check loop index range for slot assignment

   The inner loop runs from job.deadline down to 0, but slot indices start at 0, so job.deadline is out of range.

2. Step 2: Correct loop start index

   Loop should start at job.deadline - 1 to match zero-based indexing of slots array.

3. Final Answer:

   Loop index j should start from job.deadline - 1 -> Option D

4. Quick Check:

   Indexing starts at deadline-1 [OK]

Hint: Remember zero-based indexing for slots array [OK]

Common Mistakes:

• Starting loop at job.deadline causing index out of range
• Not sorting jobs by profit first
• Ignoring zero-based indexing in Go slices

5. Given jobs with deadlines and profits: {J1: deadline=4, profit=70}, {J2: deadline=1, profit=80}, {J3: deadline=1, profit=30}, {J4: deadline=2, profit=100}, {J5: deadline=3, profit=40}, what is the maximum total profit achievable by scheduling jobs before their deadlines?

hard

A. 290

B. 190

C. 230

D. 240

Solution

1. Step 1: Sort jobs by profit descending

   Order: J4(100), J2(80), J1(70), J5(40), J3(30).

2. Step 2: Schedule jobs greedily before deadlines

   J4 deadline=2 -> slot 2
   J2 deadline=1 -> slot 1
   J1 deadline=4 -> slot 4
   J5 deadline=3 -> slot 3
   J3 deadline=1 -> slot 1 taken, skip

3. Step 3: Calculate total profit

   100 + 80 + 70 + 40 = 290 (But only 4 slots available, all used)

4. Step 4: Re-check slot assignment

   Slots: 1(J2), 2(J4), 3(J5), 4(J1). Total profit = 80 + 100 + 40 + 70 = 290

5. Final Answer:

   290 -> Option A

6. Quick Check:

   Sum profits of scheduled jobs = 290 [OK]

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

Common Mistakes:

• Scheduling low profit jobs first
• Ignoring deadline constraints
• Double booking the same slot