Job Sequencing Problem
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. 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, eA Simple Solution is to 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. 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 the 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 jobstruct 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 profitbool comparison(Job a, Job b){ return (a.profit > b.profit);} // Returns minimum number of platforms requiredvoid 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 codeint 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 profitsimport 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 COdearr = [['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 CallprintJobScheduling(arr, 3) # This code is contributed# by Anubhav Raj Singh |
C#
// C# Program to find the maximum profit// job sequence from a given array// of jobs with deadlines and profits 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 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)); // Function call Console.WriteLine("Following is maximum " + "profit sequence of jobs"); Job job = new Job(); // Calling function job.printJobScheduling(arr, 3); }} // This code is contributed by avanitracchadiya2155. |
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 COdearr = [['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 CallprintJobScheduling(arr, 3) ; </script> |
Output
Following is maximum profit sequence of jobs c a e
The Time Complexity of the above solution is O(n2). It can be optimized using Priority Queue(max heap).
The algorithm goes as follow:
- Sort the jobs based on their deadlines.
- Iterate from the end and calculate the available slots between every two consecutive deadlines. Include 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.
Here is the implementation of the above algorithm.
Python3
# Program to find the maximum profit# job sequence from a given array# of jobs with deadlines and profitsimport 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 COdearr = [['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 CallprintJobScheduling(arr) # This code is contributed# by Shivam Bhagat |
Output
Following is maximum profit sequence of jobs a c e
Time complexity : O(nlog(n))
Space complexity : 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)
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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