# Weighted Job Scheduling

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

- Start Time
- Finish Time
- 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 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 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 implementation of 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

**Output: **

The optimal profit is 250

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

Below is implementation based on Dynamic Programming.

## C++

// C++ program for weighted job scheduling using Dynamic Programming. #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] 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; }

## 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

**Output: **

The optimal profit is 250

Time Complexity of the above Dynamic Programming Solution is O(n^{2}). 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