Job Sequencing Problem | Set 2 (Using Disjoint Set)

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
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
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.

1. Sort all jobs in decreasing order of profit.
2. Initialize the result sequence as first job in sorted jobs.
3. 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.

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 C++ implementation of above algorithm.

C++

```// 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 Job
struct Job
{
// Each job has id, deadline and profit
char id;
};

// A Simple Disjoint Set Data Structure
struct 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 job
bool cmp(Job a, Job b)
{
return (a.profit > b.profit);
}

// Functions returns the maximum deadline from the set
// of jobs
int findMaxDeadline(struct Job arr[], int n)
{
int ans = INT_MIN;
for (int i = 0; i < n; i++)
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

// 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)

// 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 program to test above function
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 jobs need to be "
<< "executed for maximum profit\n";
printJobScheduling(arr, n);
return 0;
return 0;
}
```

Java

```// Java program to find the maximum profit job sequence
// from a given array of jobs with deadlines and profits
import java.util.*;

// A Simple Disjoint Set Data Structure
class 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;

// Constructors
public Job() { }
{
this.id = id;
this.profit = profit;
}

// Returns the maximum deadline from the set of jobs
{
int ans = Integer.MIN_VALUE;
for (Job temp : arr)
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

// Traverse through all the jobs
for (Job temp : arr)
{
// Find the maximum available free slot for
// this job (corresponding to its 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 code
class Main
{
public static void main(String args[])
{
ArrayList<Job> arr=new ArrayList<Job>();
System.out.println("Following jobs need to be "+
"executed for maximum profit");
Job.printJobScheduling(arr);
}
}
```

Output:

```Following jobs need to be executed for maximum profit
a c e
```

This article is contributed by Chirag Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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