Job Sequencing Problem | Set 2 (Using Disjoint Set) - GeeksforGeeks

25 Jan, 2021
Given a set of n jobs where each job i has a deadline di >=1 and profit pi>=0. Only one job can be scheduled at a time. Each job takes 1 unit of time to complete. We earn the profit if and only if the job is completed by its deadline. The task is to find the subset of jobs that maximizes profit.
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 greedy solution of time complexity O(n2) is already discussed here. Below is the simple Greedy Algorithm.
Sort all jobs in decreasing order of profit.Initialize the result sequence as first job in sorted jobs.Do following for remaining n-1 jobs If the current job can fit in the current result sequence without missing the deadline, add current job to the result. Else ignore the current job.
Sort all jobs in decreasing order of profit.
Initialize the result sequence as first job in sorted jobs.
Do following for remaining n-1 jobs If the current job can fit in the current result sequence without missing the deadline, add current job to the result. Else ignore the current job.
If the current job can fit in the current result sequence without missing the deadline, add current job to the result. Else ignore the current job.
The costly operation in the Greedy solution is to assign a free slot for a job. We were traversing each and every slot for a job and assigning the greatest possible time slot(<deadline) which was available.What does greatest time slot means? Suppose that a job J1 has a deadline of time t = 5. We assign the greatest time slot which is free and less than the deadline i.e 4-5 for this job. Now another job J2 with deadline of 5 comes in, so the time slot allotted will be 3-4 since 4-5 has already been allotted to job J1.Why to assign greatest time slot(free) to a job? Now we assign the greatest possible time slot since if we assign a time slot even lesser than the available one than there might be some other job which will miss its deadline. 
Example: J1 with deadline d1 = 5, profit 40 J2 with deadline d2 = 1, profit 20 Suppose that for job J1 we assigned time slot of 0-1. Now job J2 cannot be performed since we will perform Job J1 during that time slot.Using Disjoint Set for Job Sequencing All time slots are individual sets initially. We first find the maximum deadline of all jobs. Let the max deadline be m. We create m+1 individual sets. If a job is assigned a time slot of t where t >= 0, then the job is scheduled during [t-1, t]. So a set with value X represents the time slot [X-1, X]. We need to keep track of the greatest time slot available which can be allotted to a given job having deadline. We use the parent array of Disjoint Set Data structures for this purpose. The root of the tree is always the latest available slot. If for a deadline d, there is no slot available, then root would be 0. Below are detailed steps.Initialize Disjoint Set: Creates initial disjoint sets.
// m is maximum deadline of a job
parent = new int[m + 1];

// Every node is a parent of itself
for (int i = 0; i ≤ m; i++)
    parent[i] = i;
Find : Finds the latest time slot available. 
// Returns the maximum available time slot
find(s)
{
    // Base case
    if (s == parent[s])
       return s;

    // Recursive call with path compression
    return parent[s] = find(parent[s]);
} 
Union : 
 Merges two sets.  
// Makes u as parent of v.
union(u, v)
{
   // update the greatest available
   // free slot to u
   parent[v] = u;
} 
How come find returns the latest available time slot? Initially all time slots are individual slots. So the time slot returned is always maximum. When we assign a time slot ‘t’ to a job, we do union of ‘t’ with ‘t-1’ in a way that ‘t-1’ becomes parent of ‘t’. To do this we call union(t-1, t). This means that all future queries for time slot t would now return the latest time slot available for set represented by t-1.
Implementation : The following is the implementation of above algorithm.
C++
Java
Python3
// C++ Program to find the maximum profit job sequence// from a given array of jobs with deadlines and profits#include<bits/stdc++.h>using namespace std; // A structure to represent various attributes of a Jobstruct Job{    // Each job has id, deadline and profit    char id;    int deadLine, profit;}; // A Simple Disjoint Set Data Structurestruct DisjointSet{    int *parent;     // Constructor    DisjointSet(int n)    {        parent = new int[n+1];         // Every node is a parent of itself        for (int i = 0; i <= n; i++)            parent[i] = i;    }     // Path Compression    int find(int s)    {        /* Make the parent of the nodes in the path           from u--> parent[u] point to parent[u] */        if (s == parent[s])            return s;        return parent[s] = find(parent[s]);    }     // Makes u as parent of v.    void merge(int u, int v)    {        //update the greatest available        //free slot to u        parent[v] = u;    }}; // Used to sort in descending order on the basis// of profit for each jobbool cmp(Job a, Job b){    return (a.profit > b.profit);} // Functions returns the maximum deadline from the set// of jobsint findMaxDeadline(struct Job arr[], int n){    int ans = INT_MIN;    for (int i = 0; i < n; i++)        ans = max(ans, arr[i].deadLine);    return ans;} int printJobScheduling(Job arr[], int n){    // Sort Jobs in descending order on the basis    // of their profit    sort(arr, arr + n, cmp);     // Find the maximum deadline among all jobs and    // create a disjoint set data structure with    // maxDeadline disjoint sets initially.    int maxDeadline = findMaxDeadline(arr, n);    DisjointSet ds(maxDeadline);     // Traverse through all the jobs    for (int i = 0; i < n; i++)    {        // Find the maximum available free slot for        // this job (corresponding to its deadline)        int availableSlot = ds.find(arr[i].deadLine);         // If maximum available free slot is greater        // than 0, then free slot available        if (availableSlot > 0)        {            // This slot is taken by this job 'i'            // so we need to update the greatest            // free slot. Note that, in merge, we            // make first parameter as parent of            // second parameter. So future queries            // for availableSlot will return maximum            // available slot in set of            // "availableSlot - 1"            ds.merge(ds.find(availableSlot - 1),                             availableSlot);             cout << arr[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 jobs need to be "         << "executed for maximum profit\n";    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.*; // A Simple Disjoint Set Data Structureclass DisjointSet{    int parent[];     // Constructor    DisjointSet(int n)    {        parent = new int[n + 1];         // Every node is a parent of itself        for (int i = 0; i <= n; i++)            parent[i] = i;    }     // Path Compression    int find(int s)    {        /* Make the parent of the nodes in the path           from u--> parent[u] point to parent[u] */        if (s == parent[s])            return s;        return parent[s] = find(parent[s]);    }     // Makes u as parent of v.    void merge(int u, int v)    {        //update the greatest available        //free slot to u        parent[v] = u;    }} class Job implements Comparator<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;    }     // Returns the maximum deadline from the set of jobs    public static int findMaxDeadline(ArrayList<Job> arr)    {        int ans = Integer.MIN_VALUE;        for (Job temp : arr)            ans = Math.max(temp.deadline, ans);        return ans;    }     // Prints optimal job sequence    public static void printJobScheduling(ArrayList<Job> arr)    {        // Sort Jobs in descending order on the basis        // of their profit        Collections.sort(arr, new Job());         // Find the maximum deadline among all jobs and        // create a disjoint set data structure with        // maxDeadline disjoint sets initially.        int maxDeadline = findMaxDeadline(arr);        DisjointSet dsu = new DisjointSet(maxDeadline);         // Traverse through all the jobs        for (Job temp : arr)        {            // Find the maximum available free slot for            // this job (corresponding to its deadline)            int availableSlot = dsu.find(temp.deadline);              // If maximum available free slot is greater            // than 0, then free slot available            if (availableSlot > 0)            {                // This slot is taken by this job 'i'                // so we need to update the greatest free                // slot. Note that, in merge, we make                // first parameter as parent of second                // parameter.  So future queries for                // availableSlot will return maximum slot                // from set of "availableSlot - 1"                dsu.merge(dsu.find(availableSlot - 1),                                   availableSlot);                System.out.print(temp.id + " ");            }        }        System.out.println();    }     // Used to sort in descending order on the basis    // of profit for each job    public int compare(Job j1, Job j2)    {        return j1.profit > j2.profit? -1: 1;    }} // Driver codeclass Main{    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 jobs need to be "+                           "executed for maximum profit");        Job.printJobScheduling(arr);    }}
# Python3 program to find the maximum profit# job sequence from a given array of jobs# with deadlines and profitsimport sys class DisjointSet:    def __init__(self, n):        self.parent = [i for i in range(n + 1)]     def find(self, s):             # Make the parent of nodes in the path from        # u --> parent[u] point to parent[u]        if s == self.parent[s]:            return s        self.parent[s] = self.find(self.parent[s])        return self.parent[s]     # Make us as parent of v    def merge(self, u, v):                 # Update the greatest available        # free slot to u        self.parent[v] = u def cmp(a):    return a['profit'] def findmaxdeadline(arr, n):    """    :param arr: Job array    :param n: length of array    :return: maximum deadline from the set of jobs    """    ans = - sys.maxsize - 1    for i in range(n):        ans = max(ans, arr[i]['deadline'])    return ans def printjobscheduling(arr, n):         # Sort jobs in descending order on    # basis of their profit    arr = sorted(arr, key = cmp, reverse = True)     """    Find the maximum deadline among all jobs and    create a disjoint set data structure with    max_deadline disjoint sets initially    """    max_deadline = findmaxdeadline(arr, n)    ds = DisjointSet(max_deadline)     for i in range(n):         # find maximum available free slot for        # this job (corresponding to its deadline)        available_slot = ds.find(arr[i]['deadline'])        if available_slot > 0:             # This slot is taken by this job 'i'            # so we need to update the greatest free slot.            # Note: In merge, we make first parameter            # as parent of second parameter.            # So future queries for available_slot will            # return maximum available slot in set of            # "available_slot - 1"            ds.merge(ds.find(available_slot - 1),                             available_slot)            print(arr[i]['id'], end = " ") # Driver Codeif __name__ == "__main__":    arr = [{'id': 'a', 'deadline': 2, 'profit': 100},           {'id': 'b', 'deadline': 1, 'profit': 19},           {'id': 'c', 'deadline': 2, 'profit': 27},           {'id': 'd', 'deadline': 1, 'profit': 25},           {'id': 'e', 'deadline': 3, 'profit': 15}]    n = len(arr)    print("Following jobs need to be",          "executed for maximum profit")    printjobscheduling(arr, n) # This code is contributed by Rajat Srivastava
Following jobs need to be executed for maximum profit
a c e 
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