Job Sequencing Problem

Difficulty Level : Medium
Last Updated : 19 Apr, 2021

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 the single unit of time, so the minimum possible deadline for any job is 1. How to maximize 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

A Simple Solution is to generate all subsets of given set of jobs and check individual subset for feasibility of jobs in that subset. Keep track of maximum profit among all feasible subsets. The time complexity of this solution is exponential.
This is a standard Greedy Algorithm problem.

Following is the algorithm.

1) Sort all jobs in decreasing order of profit.
2) Iterate on jobs in decreasing order of profit.For each job , do the following :
a)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.
b)If no such i exists, then ignore the job.

The Following is the implementation of above algorithm.

C++

// Program to find the maximum profit job sequence from a given array
// of jobs with deadlines and profits
#include<iostream>
#include<algorithm>
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
};
 
// This function is used for sorting all jobs according to profit
bool comparison(Job a, Job b)
{
     return (a.profit > b.profit);
}
 
// Returns minimum number of platforms reqquired
void printJobScheduling(Job arr[], int n)
{
    // Sort all jobs according to decreasing order of prfit
    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 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;
}

Java

// Program to find the maximum profit
// job sequence from a given array
// of jobs with deadlines and profits
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 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));
       
        // Function call
        System.out.println("Following is maximum "
                           + "profit sequence of jobs");
 
        Job job = new Job();
 
        // Calling function
        job.printJobScheduling(arr, 3);
    }
}
 
// This code is contributed by Prateek Gupta

Python3

# 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
 
 
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 COde
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

Javascript

<script>
 
// 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) ;
 
</script>

Output

Following is maximum profit sequence of jobs 
c a e

Time Complexity of the above solution is O(n²). It can be optimized using Disjoint Set Data Structure. Please refer below post for details.

Job Sequencing Problem | Set 2 (Using Disjoint Set)

Sources:
http://ocw.mit.edu/courses/civil-and-environmental-engineering/1-204-computer-algorithms-in-systems-engineering-spring-2010/lecture-notes/MIT1_204S10_lec10.pdf
This article is contributed by Shubham. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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