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> |
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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