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Weighted Job Scheduling

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  • Difficulty Level : Medium
  • Last Updated : 23 Mar, 2022

Given N jobs where every job is represented by following three elements of it.

  1. Start Time
  2. Finish Time
  3. Profit or Value Associated (>= 0)

Find the maximum profit subset of jobs such that no two jobs in the subset overlap. 

Example: 

Input: Number of Jobs n = 4
       Job Details {Start Time, Finish Time, Profit}
       Job 1:  {1, 2, 50} 
       Job 2:  {3, 5, 20}
       Job 3:  {6, 19, 100}
       Job 4:  {2, 100, 200}
Output: The maximum profit is 250.
We can get the maximum profit by scheduling jobs 1 and 4.
Note that there is longer schedules possible Jobs 1, 2 and 3 
but the profit with this schedule is 20+50+100 which is less than 250.

A simple version of this problem is discussed here where every job has the same profit or value. The Greedy Strategy for activity selection doesn’t work here as a schedule with more jobs may have smaller profit or value.

The above problem can be solved using the following recursive solution.  

1) First sort jobs according to finish time.
2) Now apply following recursive process. 
   // Here arr[] is array of n jobs
   findMaximumProfit(arr[], n)
   {
     a) if (n == 1) return arr[0];
     b) Return the maximum of following two profits.
         (i) Maximum profit by excluding current job, i.e., 
             findMaximumProfit(arr, n-1)
         (ii) Maximum profit by including the current job            
   }

How to find the profit including current job?
The idea is to find the latest job before the current job (in 
sorted array) that doesn't conflict with current job 'arr[n-1]'. 
Once we find such a job, we recur for all jobs till that job and
add profit of current job to result.
In the above example, "job 1" is the latest non-conflicting
for "job 4" and "job 2" is the latest non-conflicting for "job 3".

The following is the implementation of the above naive recursive method. 

C++




// C++ program for weighted job scheduling using Naive Recursive Method
#include <iostream>
#include <algorithm>
using namespace std;
 
// A job has start time, finish time and profit.
struct Job
{
    int start, finish, profit;
};
 
// A utility function that is used for sorting events
// according to finish time
bool jobComparator(Job s1, Job s2)
{
    return (s1.finish < s2.finish);
}
 
// Find the latest job (in sorted array) that doesn't
// conflict with the job[i]. If there is no compatible job,
// then it returns -1.
int latestNonConflict(Job arr[], int i)
{
    for (int j=i-1; j>=0; j--)
    {
        if (arr[j].finish <= arr[i-1].start)
            return j;
    }
    return -1;
}
 
// A recursive function that returns the maximum possible
// profit from given array of jobs.  The array of jobs must
// be sorted according to finish time.
int findMaxProfitRec(Job arr[], int n)
{
    // Base case
    if (n == 1) return arr[n-1].profit;
 
    // Find profit when current job is included
    int inclProf = arr[n-1].profit;
    int i = latestNonConflict(arr, n);
    if (i != -1)
      inclProf += findMaxProfitRec(arr, i+1);
 
    // Find profit when current job is excluded
    int exclProf = findMaxProfitRec(arr, n-1);
 
    return max(inclProf,  exclProf);
}
 
// The main function that returns the maximum possible
// profit from given array of jobs
int findMaxProfit(Job arr[], int n)
{
    // Sort jobs according to finish time
    sort(arr, arr+n, jobComparator);
 
    return findMaxProfitRec(arr, n);
}
 
// Driver program
int main()
{
    Job arr[] = {{3, 10, 20}, {1, 2, 50}, {6, 19, 100}, {2, 100, 200}};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << "The optimal profit is " << findMaxProfit(arr, n);
    return 0;
}

Java




// JAVA program for weighted job scheduling using Naive Recursive Method
import java.util.*;
class GFG
{
   
// A job has start time, finish time and profit.
static class Job
{
    int start, finish, profit;
    Job(int start, int finish, int profit)
     {
        this.start = start;
        this.finish = finish;
        this.profit = profit;
     }
}
 
// Find the latest job (in sorted array) that doesn't
// conflict with the job[i]. If there is no compatible job,
// then it returns -1.
static int latestNonConflict(Job arr[], int i)
{
    for (int j = i - 1; j >= 0; j--)
    {
        if (arr[j].finish <= arr[i - 1].start)
            return j;
    }
    return -1;
}
 
// A recursive function that returns the maximum possible
// profit from given array of jobs. The array of jobs must
// be sorted according to finish time.
static int findMaxProfitRec(Job arr[], int n)
{
    // Base case
    if (n == 1) return arr[n-1].profit;
 
    // Find profit when current job is included
    int inclProf = arr[n-1].profit;
    int i = latestNonConflict(arr, n);
    if (i != -1)
    inclProf += findMaxProfitRec(arr, i+1);
 
    // Find profit when current job is excluded
    int exclProf = findMaxProfitRec(arr, n-1);
 
    return Math.max(inclProf, exclProf);
}
 
// The main function that returns the maximum possible
// profit from given array of jobs
static int findMaxProfit(Job arr[], int n)
{
    // Sort jobs according to finish time
    Arrays.sort(arr,new Comparator<Job>(){
       public int compare(Job j1,Job j2)
        {
           return j1.finish-j2.finish;
        }
       });
 
    return findMaxProfitRec(arr, n);
}
 
// Driver program
public static void main(String args[])
{
   int m = 4;
   Job arr[] = new Job[m];
    arr[0] = new Job(3, 10, 20);
    arr[1] = new Job(1, 2, 50);
    arr[2] = new Job(6, 19, 100);
    arr[3] = new Job(2, 100, 200);
    int n =arr.length;
    System.out.println("The optimal profit is " + findMaxProfit(arr, n));
}
}
 
// This code is contributed by Debojyoti Mandal

Python3




# Python3 program for weighted job scheduling using
# Naive Recursive Method
 
# Importing the following module to sort array
# based on our custom comparison function
from functools import cmp_to_key
 
# A job has start time, finish time and profit
class Job:
     
    def __init__(self, start, finish, profit):
         
        self.start = start
        self.finish = finish
        self.profit = profit
 
# A utility function that is used for
# sorting events according to finish time
def jobComparator(s1, s2):
     
    return s1.finish < s2.finish
 
# Find the latest job (in sorted array) that
# doesn't conflict with the job[i]. If there
# is no compatible job, then it returns -1
def latestNonConflict(arr, i):
     
    for j in range(i - 1, -1, -1):
        if arr[j].finish <= arr[i - 1].start:
            return j
             
    return -1
 
# A recursive function that returns the
# maximum possible profit from given
# array of jobs. The array of jobs must
# be sorted according to finish time
def findMaxProfitRec(arr, n):
     
    # Base case
    if n == 1:
        return arr[n - 1].profit
 
    # Find profit when current job is included
    inclProf = arr[n - 1].profit
    i = latestNonConflict(arr, n)
     
    if i != -1:
        inclProf += findMaxProfitRec(arr, i + 1)
 
    # Find profit when current job is excluded
    exclProf = findMaxProfitRec(arr, n - 1)
    return max(inclProf, exclProf)
 
# The main function that returns the maximum
# possible profit from given array of jobs
def findMaxProfit(arr, n):
     
    # Sort jobs according to finish time
    arr = sorted(arr, key = cmp_to_key(jobComparator))
    return findMaxProfitRec(arr, n)
 
# Driver code
values = [ (3, 10, 20), (1, 2, 50),
           (6, 19, 100), (2, 100, 200) ]
arr = []
for i in values:
    arr.append(Job(i[0], i[1], i[2]))
     
n = len(arr)
 
print("The optimal profit is", findMaxProfit(arr, n))
 
# This code is code contributed by Kevin Joshi

Javascript




<script>
 
// JavaScript program for weighted job scheduling using
// Naive Recursive Method
 
// A job has start time, finish time and profit
class Job
{
    constructor(start, finish, profit)
    {
        this.start = start
        this.finish = finish
        this.profit = profit
    }
}
 
// A utility function that is used for
// sorting events according to finish time
function jobComparator(s1, s2){
     
    return s2.finish - s1.finish;
}
 
// Find the latest job (in sorted array) that
// doesn't conflict with the job[i]. If there
// is no compatible job, then it returns -1
function latestNonConflict(arr, i){
     
    for(let j = i - 1; j >= 0; j--)
    {
        if(arr[j].finish <= arr[i - 1].start)
            return j
    }       
    return -1
}
 
// A recursive function that returns the
// maximum possible profit from given
// array of jobs. The array of jobs must
// be sorted according to finish time
function findMaxProfitRec(arr, n){
     
    // Base case
    if(n == 1)
        return arr[n - 1].profit
 
    // Find profit when current job is included
    let inclProf = arr[n - 1].profit
    let i = latestNonConflict(arr, n)
     
    if(i != -1)
        inclProf += findMaxProfitRec(arr, i + 1)
 
    // Find profit when current job is excluded
    let exclProf = findMaxProfitRec(arr, n - 1)
    return Math.max(inclProf, exclProf)
}
 
// The main function that returns the maximum
// possible profit from given array of jobs
function findMaxProfit(arr, n){
     
    // Sort jobs according to finish time
    arr.sort(jobComparator)
    return findMaxProfitRec(arr, n)
 
}
 
// Driver code
let values = [ [3, 10, 20], [1, 2, 50],
           [6, 19, 100], [2, 100, 200] ]
let arr = []
for(let i of values)
    arr.push(new Job(i[0], i[1], i[2]))
     
let n = arr.length
document.write("The optimal profit is ", findMaxProfit(arr, n),"</br>")
 
// This code is code contributed by shinjanpatra
 
</script>

Output: 

The optimal profit is 250

The above solution may contain many overlapping subproblems. For example, if lastNonConflicting() always returns the previous job, then findMaxProfitRec(arr, n-1) is called twice and the time complexity becomes O(n*2n). As another example when lastNonConflicting() returns previous to the previous job, there are two recursive calls, for n-2 and n-1. In this example case, recursion becomes the same as Fibonacci Numbers. 

So this problem has both properties of Dynamic Programming, Optimal Substructure, and Overlapping Subproblems
Like other Dynamic Programming Problems, we can solve this problem by making a table that stores solutions of subproblems.

Below is an implementation based on Dynamic Programming. 

C++




// C++ program for weighted job scheduling using Dynamic
// Programming.
#include <algorithm>
#include <iostream>
using namespace std;
 
// A job has start time, finish time and profit.
struct Job {
    int start, finish, profit;
};
 
// A utility function that is used for sorting events
// according to finish time
bool jobComparator(Job s1, Job s2)
{
    return (s1.finish < s2.finish);
}
 
// Find the latest job (in sorted array) that doesn't
// conflict with the job[i]
int latestNonConflict(Job arr[], int i)
{
    for (int j = i - 1; j >= 0; j--) {
        if (arr[j].finish <= arr[i].start)
            return j;
    }
    return -1;
}
 
// The main function that returns the maximum possible
// profit from given array of jobs
int findMaxProfit(Job arr[], int n)
{
    // Sort jobs according to finish time
    sort(arr, arr + n, jobComparator);
 
    // Create an array to store solutions of subproblems.
    // table[i] stores the profit for jobs till arr[i]
    // (including arr[i])
    int* table = new int[n];
    table[0] = arr[0].profit;
 
    // Fill entries in M[] using recursive property
    for (int i = 1; i < n; i++) {
        // Find profit including the current job
        int inclProf = arr[i].profit;
        int l = latestNonConflict(arr, i);
        if (l != -1)
            inclProf += table[l];
 
        // Store maximum of including and excluding
        table[i] = max(inclProf, table[i - 1]);
    }
 
    // Store result and free dynamic memory allocated for
    // table[]
    int result = table[n - 1];
    delete[] table;
 
    return result;
}
 
// Driver program
int main()
{
    Job arr[] = { { 3, 10, 20 },
                  { 1, 2, 50 },
                  { 6, 19, 100 },
                  { 2, 100, 200 } };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "The optimal profit is "
         << findMaxProfit(arr, n);
    return 0;
}

Java




// JAVA program for weighted job scheduling using Naive
// Recursive Method
import java.util.*;
class GFG {
 
    // A job has start time, finish time and profit.
    static class Job {
        int start, finish, profit;
        Job(int start, int finish, int profit)
        {
            this.start = start;
            this.finish = finish;
            this.profit = profit;
        }
    }
 
    // Find the latest job (in sorted array) that doesn't
    // conflict with the job[i]. If there is no compatible
    // job, then it returns -1.
    static int latestNonConflict(Job arr[], int i)
    {
        for (int j = i - 1; j >= 0; j--) {
            // finish before next is started
            if (arr[j].finish <= arr[i - 1].start)
                return j;
        }
        return -1;
    }
 
    static int findMaxProfitDP(Job arr[], int n)
    {
 
        // Create an array to store solutions of
        // subproblems.  table[i] stores the profit for jobs
        // till arr[i] (including arr[i])
        int[] table = new int[n];
        table[0] = arr[0].profit;
 
        // Fill entries in M[] using recursive property
        for (int i = 1; i < n; i++) {
            // Find profit including the current job
            int inclProf = arr[i].profit;
            int l = latestNonConflict(arr, i);
            if (l != -1)
                inclProf += table[l];
 
            // Store maximum of including and excluding
            table[i] = Math.max(inclProf, table[i - 1]);
        }
 
        // Store result and free dynamic memory allocated
        // for table[]
        int result = table[n - 1];
 
        return result;
    }
 
    // The main function that returns the maximum possible
    // profit from given array of jobs
    static int findMaxProfit(Job arr[], int n)
    {
        // Sort jobs according to finish time
        Arrays.sort(arr, new Comparator<Job>() {
            public int compare(Job j1, Job j2)
            {
                return j1.finish - j2.finish;
            }
        });
 
        return findMaxProfitDP(arr, n);
    }
 
    // Driver program
    public static void main(String args[])
    {
        int m = 4;
        Job arr[] = new Job[m];
        arr[0] = new Job(3, 10, 20);
        arr[1] = new Job(1, 2, 50);
        arr[2] = new Job(6, 19, 100);
        arr[3] = new Job(2, 100, 200);
        int n = arr.length;
        System.out.println("The optimal profit is "
                           + findMaxProfit(arr, n));
    }
}

Python3




# Python3 program for weighted job scheduling
# using Dynamic Programming
 
# Importing the following module to sort array
# based on our custom comparison function
from functools import cmp_to_key
 
# A job has start time, finish time and profit
 
 
class Job:
 
    def __init__(self, start, finish, profit):
 
        self.start = start
        self.finish = finish
        self.profit = profit
 
# A utility function that is used for sorting
# events according to finish time
 
 
def jobComparator(s1, s2):
 
    return s1.finish < s2.finish
 
# Find the latest job (in sorted array) that
# doesn't conflict with the job[i]. If there
# is no compatible job, then it returns -1
 
 
def latestNonConflict(arr, i):
 
    for j in range(i - 1, -1, -1):
        if arr[j].finish <= arr[i - 1].start:
            return j
 
    return -1
 
# The main function that returns the maximum possible
# profit from given array of jobs
 
 
def findMaxProfit(arr, n):
 
    # Sort jobs according to finish time
    arr = sorted(arr, key=cmp_to_key(jobComparator))
 
    # Create an array to store solutions of subproblems.
    # table[i] stores the profit for jobs till arr[i]
    # (including arr[i])
    table = [None] * n
    table[0] = arr[0].profit
 
    # Fill entries in M[] using recursive property
    for i in range(1, n):
 
        # Find profit including the current job
        inclProf = arr[i].profit
        l = latestNonConflict(arr, i)
 
        if l != -1:
            inclProf += table[l]
 
        # Store maximum of including and excluding
        table[i] = max(inclProf, table[i - 1])
 
    # Store result and free dynamic memory
    # allocated for table[]
    result = table[n - 1]
 
    return result
 
 
# Driver code
values = [(3, 10, 20), (1, 2, 50),
          (6, 19, 100), (2, 100, 200)]
arr = []
for i in values:
    arr.append(Job(i[0], i[1], i[2]))
 
n = len(arr)
 
print("The optimal profit is", findMaxProfit(arr, n))
 
# This code is contributed by Kevin Joshi

Javascript




<script>
 
// JavaScript program for weighted job scheduling using
// Naive Recursive Method
 
// A job has start time, finish time and profit
class Job{
     
    constructor(start, finish, profit){
         
        this.start = start
        this.finish = finish
        this.profit = profit
    }
 
}
 
// A utility function that is used for
// sorting events according to finish time
function jobComparator(s1, s2){
     
    return s1.finish - s2.finish
}
 
// Find the latest job (in sorted array) that
// doesn't conflict with the job[i]. If there
// is no compatible job, then it returns -1
function latestNonConflict(arr, i){
     
    for(let j=i-1;j>=0;j--){
        if (arr[j].finish <= arr[i - 1].start)
            return j
    }
             
    return -1
}
 
 
// The main function that returns the maximum
// possible profit from given array of jobs
function findMaxProfit(arr, n){
     
    // Sort jobs according to finish time
    arr.sort(jobComparator)
 
    // Create an array to store solutions of subproblems.
    // table[i] stores the profit for jobs till arr[i]
    // (including arr[i])
    let table = new Array(n).fill(null)
    table[0] = arr[0].profit
 
    // Fill entries in M[] using recursive property
    for(let i=1;i<n;i++){
 
        // Find profit including the current job
        let inclProf = arr[i].profit
        let l = latestNonConflict(arr, i)
 
        if (l != -1)
            inclProf += table[l]
 
        // Store maximum of including and excluding
        table[i] = Math.max(inclProf, table[i - 1])
     
    }
 
    // Store result and free dynamic memory
    // allocated for table[]
    let result = table[n - 1]
 
    return result
}
 
// Driver code
let values = [ [3, 10, 20], [1, 2, 50], [6, 19, 100], [2, 100, 200] ]
let arr = []
for(let i of values)
    arr.push(new Job(i[0], i[1], i[2]))
     
let n = arr.length
 
document.write("The optimal profit is ", findMaxProfit(arr, n),"</br>")
 
// This code is code contributed by shinjanpatra
 
</script>

Output: 

The optimal profit is 250

Time Complexity of the above Dynamic Programming Solution is O(n2). Note that the above solution can be optimized to O(nLogn) using Binary Search in latestNonConflict() instead of linear search. Thanks to Garvit for suggesting this optimization. Please refer below post for details.

Weighted Job Scheduling in O(n Log n) time

References: 
http://courses.cs.washington.edu/courses/cse521/13wi/slides/06dp-sched.pdf

This article is contributed by Shivam. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 


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