Skip to content
Related Articles

Related Articles

Job Sequencing Problem

View Discussion
Improve Article
Save Article
  • Difficulty Level : Medium
  • Last Updated : 22 Aug, 2022
View Discussion
Improve Article
Save Article

Given an array of jobs where every job has a deadline and associated profit if the job is finished before the deadline. It is also given that every job takes a single unit of time, so the minimum possible deadline for any job is 1. Maximize the total profit if only one job can be scheduled at a time.

Examples: 

Input: Four Jobs with following deadlines and profits

JobID  Deadline  Profit

  a           4          20   
  b           1          10
  c           1          40  
  d          1          30

Output: Following is maximum profit sequence of jobs: c, a   

Input:  Five Jobs with following deadlines and profits

JobID   Deadline  Profit

  a            2          100
  b            1          19
  c            2          27
 d            1          25
 e            3          15

Output: Following is maximum profit sequence of jobs: c, a, e

Naive Approach: To solve the problem follow the below idea:

Generate all subsets of a given set of jobs and check individual subsets for the feasibility of jobs in that subset. Keep track of maximum profit among all feasible subsets.

Greedy approach for job sequencing problem:

Greedily choose the jobs with maximum profit first, by sorting the jobs in decreasing order of their profit. This would help to maximize the total profit as choosing the job with maximum profit for every time slot will eventually maximize the total profit

Follow the given steps to solve the problem:

  • Sort all jobs in decreasing order of profit. 
  • Iterate on jobs in decreasing order of profit.For each job , do the following : 
    • Find a time slot i, such that slot is empty and i < deadline and i is greatest.Put the job in 
      this slot and mark this slot filled. 
    • If no such i exists, then ignore the job. 

Below is the implementation of the above approach:

C




// C program for the above approach
 
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
 
// A structure to represent a job
typedef struct Job {
   
    char id; // Job Id
    int dead; // Deadline of job
    int profit; // Profit if job is over before or on
                // deadline
} Job;
 
// This function is used for sorting all jobs according to
// profit
int compare(const void* a, const void* b)
{
    Job* temp1 = (Job*)a;
    Job* temp2 = (Job*)b;
    return (temp2->profit - temp1->profit);
}
 
// Find minimum between two numbers.
int min(int num1, int num2)
{
    return (num1 > num2) ? num2 : num1;
}
 
// Returns maximum profit from jobs
void printJobScheduling(Job arr[], int n)
{
    // Sort all jobs according to decreasing order of profit
    qsort(arr, n, sizeof(Job), compare);
 
    int result[n]; // To store result (Sequence of jobs)
    bool slot[n]; // To keep track of free time slots
 
    // Initialize all slots to be free
    for (int i = 0; i < n; i++)
        slot[i] = false;
 
    // Iterate through all given jobs
    for (int i = 0; i < n; i++) {
       
        // Find a free slot for this job (Note that we start
        // from the last possible slot)
        for (int j = min(n, arr[i].dead) - 1; j >= 0; j--) {
           
            // Free slot found
            if (slot[j] == false) {
                result[j] = i; // Add this job to result
                slot[j] = true; // Make this slot occupied
                break;
            }
        }
    }
 
    // Print the result
    for (int i = 0; i < n; i++)
        if (slot[i])
            printf("%c ", arr[result[i]].id);
}
 
// Driver's code
int main()
{
    Job arr[] = { { 'a', 2, 100 },
                  { 'b', 1, 19 },
                  { 'c', 2, 27 },
                  { 'd', 1, 25 },
                  { 'e', 3, 15 } };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf(
        "Following is maximum profit sequence of jobs \n");
 
    // Function call
    printJobScheduling(arr, n);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

C++




// C++ code for the above approach
 
#include <algorithm>
#include <iostream>
using namespace std;
 
// A structure to represent a job
struct Job {
   
    char id; // Job Id
    int dead; // Deadline of job
    int profit; // Profit if job is over before or on
                // deadline
};
 
// Comparator function for sorting jobs
bool comparison(Job a, Job b)
{
    return (a.profit > b.profit);
}
 
// Returns maximum profit from jobs
void printJobScheduling(Job arr[], int n)
{
    // Sort all jobs according to decreasing order of profit
    sort(arr, arr + n, comparison);
 
    int result[n]; // To store result (Sequence of jobs)
    bool slot[n]; // To keep track of free time slots
 
    // Initialize all slots to be free
    for (int i = 0; i < n; i++)
        slot[i] = false;
 
    // Iterate through all given jobs
    for (int i = 0; i < n; i++) {
        // Find a free slot for this job (Note that we start
        // from the last possible slot)
        for (int j = min(n, arr[i].dead) - 1; j >= 0; j--) {
            // Free slot found
            if (slot[j] == false) {
                result[j] = i; // Add this job to result
                slot[j] = true; // Make this slot occupied
                break;
            }
        }
    }
 
    // Print the result
    for (int i = 0; i < n; i++)
        if (slot[i])
            cout << arr[result[i]].id << " ";
}
 
// Driver's code
int main()
{
    Job arr[] = { { 'a', 2, 100 },
                  { 'b', 1, 19 },
                  { 'c', 2, 27 },
                  { 'd', 1, 25 },
                  { 'e', 3, 15 } };
   
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Following is maximum profit sequence of jobs "
            "\n";
 
    // Function call
    printJobScheduling(arr, n);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

Java




// Java code for the above approach
 
import java.util.*;
 
class Job {
   
    // Each job has a unique-id,profit and deadline
    char id;
    int deadline, profit;
 
    // Constructors
    public Job() {}
 
    public Job(char id, int deadline, int profit)
    {
        this.id = id;
        this.deadline = deadline;
        this.profit = profit;
    }
 
    // Function to schedule the jobs take 2 arguments
    // arraylist and no of jobs to schedule
    void printJobScheduling(ArrayList<Job> arr, int t)
    {
        // Length of array
        int n = arr.size();
       
        // Sort all jobs according to decreasing order of
        // profit
        Collections.sort(arr,
                         (a, b) -> b.profit - a.profit);
 
        // To keep track of free time slots
        boolean result[] = new boolean[t];
 
        // To store result (Sequence of jobs)
        char job[] = new char[t];
 
        // Iterate through all given jobs
        for (int i = 0; i < n; i++) {
            // Find a free slot for this job (Note that we
            // start from the last possible slot)
            for (int j
                 = Math.min(t - 1, arr.get(i).deadline - 1);
                 j >= 0; j--) {
                // Free slot found
                if (result[j] == false) {
                    result[j] = true;
                    job[j] = arr.get(i).id;
                    break;
                }
            }
        }
 
        // Print the sequence
        for (char jb : job)
            System.out.print(jb + " ");
        System.out.println();
    }
 
    // Driver's code
    public static void main(String args[])
    {
        ArrayList<Job> arr = new ArrayList<Job>();
        arr.add(new Job('a', 2, 100));
        arr.add(new Job('b', 1, 19));
        arr.add(new Job('c', 2, 27));
        arr.add(new Job('d', 1, 25));
        arr.add(new Job('e', 3, 15));
 
        System.out.println(
            "Following is maximum profit sequence of jobs");
 
        Job job = new Job();
 
        // Function call
        job.printJobScheduling(arr, 3);
    }
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

Python3




# Python3 code for the above approach
 
# function to schedule the jobs take 2
# arguments array and no of jobs to schedule
 
 
def printJobScheduling(arr, t):
 
    # length of array
    n = len(arr)
 
    # Sort all jobs according to
    # decreasing order of profit
    for i in range(n):
        for j in range(n - 1 - i):
            if arr[j][2] < arr[j + 1][2]:
                arr[j], arr[j + 1] = arr[j + 1], arr[j]
 
    # To keep track of free time slots
    result = [False] * t
 
    # To store result (Sequence of jobs)
    job = ['-1'] * t
 
    # Iterate through all given jobs
    for i in range(len(arr)):
 
        # Find a free slot for this job
        # (Note that we start from the
        # last possible slot)
        for j in range(min(t - 1, arr[i][1] - 1), -1, -1):
 
            # Free slot found
            if result[j] is False:
                result[j] = True
                job[j] = arr[i][0]
                break
 
    # print the sequence
    print(job)
 
 
# Driver's Code
if __name__ == '__main__':
    arr = [['a', 2, 100],  # Job Array
              ['b', 1, 19],
              ['c', 2, 27],
              ['d', 1, 25],
              ['e', 3, 15]]
 
 
    print("Following is maximum profit sequence of jobs")
 
    # Function Call
    printJobScheduling(arr, 3)
 
# This code is contributed
# by Anubhav Raj Singh

C#




// C# Program for the above approach
 
using System;
using System.Collections.Generic;
 
class GFG : IComparer<Job> {
    public int Compare(Job x, Job y)
    {
        if (x.profit == 0 || y.profit == 0) {
            return 0;
        }
 
        // CompareTo() method
        return (y.profit).CompareTo(x.profit);
    }
}
 
public class Job {
 
    // Each job has a unique-id,
    // profit and deadline
    char id;
    public int deadline, profit;
 
    // Constructors
    public Job() {}
 
    public Job(char id, int deadline, int profit)
    {
        this.id = id;
        this.deadline = deadline;
        this.profit = profit;
    }
 
    // Function to schedule the jobs take 2
    // arguments arraylist and no of jobs to schedule
    void printJobScheduling(List<Job> arr, int t)
    {
        // Length of array
        int n = arr.Count;
 
        GFG gg = new GFG();
        // Sort all jobs according to
        // decreasing order of profit
        arr.Sort(gg);
 
        // To keep track of free time slots
        bool[] result = new bool[t];
 
        // To store result (Sequence of jobs)
        char[] job = new char[t];
 
        // Iterate through all given jobs
        for (int i = 0; i < n; i++) {
            // Find a free slot for this job
            // (Note that we start from the
            // last possible slot)
            for (int j
                 = Math.Min(t - 1, arr[i].deadline - 1);
                 j >= 0; j--) {
 
                // Free slot found
                if (result[j] == false) {
                    result[j] = true;
                    job[j] = arr[i].id;
                    break;
                }
            }
        }
 
        // Print the sequence
        foreach(char jb in job) { Console.Write(jb + " "); }
        Console.WriteLine();
    }
 
    // Driver's code
    static public void Main()
    {
 
        List<Job> arr = new List<Job>();
 
        arr.Add(new Job('a', 2, 100));
        arr.Add(new Job('b', 1, 19));
        arr.Add(new Job('c', 2, 27));
        arr.Add(new Job('d', 1, 25));
        arr.Add(new Job('e', 3, 15));
 
        Console.WriteLine("Following is maximum "
                          + "profit sequence of jobs");
 
        Job job = new Job();
 
        // Function call
        job.printJobScheduling(arr, 3);
    }
}
 
// This code is contributed by avanitracchadiya2155.

Javascript




// Program to find the maximum profit
// job sequence from a given array
// of jobs with deadlines and profits
 
// function to schedule the jobs take 2
// arguments array and no of jobs to schedule
 
function printJobScheduling(arr, t){
    // length of array
    let n = arr.length;
 
    // Sort all jobs according to
    // decreasing order of profit
    for(let i=0;i<n;i++){
        for(let j = 0;j<(n - 1 - i);j++){
            if(arr[j][2] < arr[j + 1][2]){
                let temp = arr[j];
                arr[j] = arr[j + 1];
                arr[j + 1] = temp;
            }
         }
     }
 
    // To keep track of free time slots
    let result = [];
 
    // To store result (Sequence of jobs)
    let job = [];
    for(let i = 0;i<t;i++){
        job[i] = '-1';
        result[i] = false;
    }
 
    // Iterate through all given jobs
    for(let i= 0;i<arr.length;i++){
        // Find a free slot for this job
        // (Note that we start from the
        // last possible slot)
        for(let j = (t - 1, arr[i][1] - 1);j>=0;j--){
            // Free slot found
            if(result[j] == false){
                result[j] = true;
                job[j] = arr[i][0];
                break;
            }
        }
    }
 
    // print the sequence
    document.write(job);
}
 
// Driver COde
arr = [['a', 2, 100],  // Job Array
       ['b', 1, 19],
       ['c', 2, 27],
       ['d', 1, 25],
       ['e', 3, 15]];
 
document.write("Following is maximum profit sequence of jobs ");
document.write("<br>");
 
// Function Call
printJobScheduling(arr, 3) ;

Output

Following is maximum profit sequence of jobs 
c a e 

Time Complexity: O(N2)
Auxiliary Space: O(N)

Job sequencing problem using Priority-Queue (Max-Heap):

Sort the jobs in the increasing order of their deadlines and then calculate the available slots between every two consecutive deadlines while iterating from the end. Include the profit of the job at the root of the Max-Heap while the empty slots are available and Heap is not empty, as this would help to choose the jobs with maximum profit for every set of available slots.

Follow the given steps to solve the problem:

  • Sort the jobs based on their deadlines.
  • Iterate from the end and calculate the available slots between every two consecutive deadlines. Insert the profit, deadline, and job ID of ith job in the max heap.
  • While the slots are available and there are jobs left in the max heap, include the job ID with maximum profit and deadline in the result.
  • Sort the result array based on their deadlines.

Below is the implementation of the above approach:

C++




// C++ code for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// A structure to represent a job
struct Job {
   
    char id; // Job Id
    int dead; // Deadline of job
    int profit; // Profit earned if job is completed before
                // deadline
};
 
// Custom sorting helper struct which is used for sorting
// all jobs according to profit
struct jobProfit {
    bool operator()(Job const& a, Job const& b)
    {
        return (a.profit < b.profit);
    }
};
 
// Returns maximum profit from jobs
void printJobScheduling(Job arr[], int n)
{
    vector<Job> result;
    sort(arr, arr + n,
         [](Job a, Job b) { return a.dead < b.dead; });
   
    // set a custom priority queue
    priority_queue<Job, vector<Job>, jobProfit> pq;
   
    for (int i = n - 1; i >= 0; i--) {
        int slot_available;
       
        // we count the slots available between two jobs
        if (i == 0) {
            slot_available = arr[i].dead;
        }
        else {
            slot_available = arr[i].dead - arr[i - 1].dead;
        }
       
        // include the profit of job(as priority),
        // deadline and job_id in maxHeap
        pq.push(arr[i]);
       
        while (slot_available > 0 && pq.size() > 0) {
           
            // get the job with the most profit
            Job job = pq.top();
            pq.pop();
           
            // reduce the slots
            slot_available--;
           
            // add it to the answer
            result.push_back(job);
        }
    }
   
    // sort the result based on the deadline
    sort(result.begin(), result.end(),
         [&](Job a, Job b) { return a.dead < b.dead; });
   
    // print the result
    for (int i = 0; i < result.size(); i++)
        cout << result[i].id << ' ';
    cout << endl;
}
 
// Driver's code
int main()
{
    Job arr[] = { { 'a', 2, 100 },
                  { 'b', 1, 19 },
                  { 'c', 2, 27 },
                  { 'd', 1, 25 },
                  { 'e', 3, 15 } };
   
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Following is maximum profit sequence of jobs "
            "\n";
 
    // Function call
    printJobScheduling(arr, n);
    return 0;
}
 
// This code is contributed By Reetu Raj Dubey

Java




// Java implementation of above approach
 
// Program to find the maximum profit
// job sequence from a given array
// of jobs with deadlines and profits
import java.util.*;
 
class GFG {
 
    // a class to represent job
    static class Job {
        char job_id;
        int deadline;
        int profit;
        Job(char job_id, int deadline, int profit)
        {
            this.deadline = deadline;
            this.job_id = job_id;
            this.profit = profit;
        }
    }
 
    static void printJobScheduling(ArrayList<Job> arr)
    {
        int n = arr.size();
 
        // sorting the array on the
        // basis of their deadlines
        Collections.sort(arr, (a, b) -> {
            return a.deadline - b.deadline;
        });
 
        // initialise the result array and maxHeap
        ArrayList<Job> result = new ArrayList<>();
        PriorityQueue<Job> maxHeap = new PriorityQueue<>(
            (a, b) -> { return b.profit - a.profit; });
 
        // starting the iteration from the end
        for (int i = n - 1; i > -1; i--) {
            int slot_available;
           
            // calculate slots between two deadlines
            if (i == 0) {
                slot_available = arr.get(i).deadline;
            }
            else {
                slot_available = arr.get(i).deadline
                                 - arr.get(i - 1).deadline;
            }
 
            // include the profit of job(as priority),
            // deadline and job_id in maxHeap
            maxHeap.add(arr.get(i));
 
            while (slot_available > 0
                   && maxHeap.size() > 0) {
 
                // get the job with max_profit
                Job job = maxHeap.remove();
 
                // reduce the slots
                slot_available--;
 
                // include the job in the result array
                result.add(job);
            }
        }
 
        // jobs included might be shuffled
        // sort the result array by their deadlines
        Collections.sort(result, (a, b) -> {
            return a.deadline - b.deadline;
        });
       
        for (Job job : result) {
            System.out.print(job.job_id + " ");
        }
       
        System.out.println();
    }
 
    // Driver's Code
    public static void main(String[] args)
    {
        ArrayList<Job> arr = new ArrayList<Job>();
 
        arr.add(new Job('a', 2, 100));
        arr.add(new Job('b', 1, 19));
        arr.add(new Job('c', 2, 27));
        arr.add(new Job('d', 1, 25));
        arr.add(new Job('e', 3, 15));
       
        System.out.println("Following is maximum "
                           + "profit sequence of jobs");
 
        // Function call
        printJobScheduling(arr);
    }
}
 
// This code is contributed by Karandeep Singh

Python3




# Python3 program for the above approach
import heapq
 
 
def printJobScheduling(arr):
    n = len(arr)
 
    # arr[i][0] = job_id, arr[i][1] = deadline, arr[i][2] = profit
 
    # sorting the array on the
    # basis of their deadlines
    arr.sort(key=lambda x: x[1])
 
    # initialise the result array and maxHeap
    result = []
    maxHeap = []
 
    # starting the iteration from the end
    for i in range(n - 1, -1, -1):
 
        # calculate slots between two deadlines
        if i == 0:
            slots_available = arr[i][1]
        else:
            slots_available = arr[i][1] - arr[i - 1][1]
 
        # include the profit of job(as priority), deadline
        # and job_id in maxHeap
        # note we push negative value in maxHeap to convert
        # min heap to max heap in python
        heapq.heappush(maxHeap, (-arr[i][2], arr[i][1], arr[i][0]))
 
        while slots_available and maxHeap:
 
            # get the job with max_profit
            profit, deadline, job_id = heapq.heappop(maxHeap)
 
            # reduce the slots
            slots_available -= 1
 
            # include the job in the result array
            result.append([job_id, deadline])
 
    # jobs included might be shuffled
    # sort the result array by their deadlines
    result.sort(key=lambda x: x[1])
 
    for job in result:
        print(job[0], end=" ")
    print()
 
 
# Driver's Code
if __name__ == '__main__':
    arr = [['a', 2, 100],  # Job Array
           ['b', 1, 19],
           ['c', 2, 27],
           ['d', 1, 25],
           ['e', 3, 15]]
 
    print("Following is maximum profit sequence of jobs")
 
    # Function Call
    printJobScheduling(arr)
 
# This code is contributed
# by Shivam Bhagat

Output

Following is maximum profit sequence of jobs
a c e 

Time Complexity: O(N log N)
Auxiliary Space: O(N)

It can also be optimized using Disjoint Set Data Structure. Please refer to the below post for details.
Job Sequencing Problem | Set 2 (Using Disjoint Set)
 

This article is contributed by Shubham. Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!