# 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

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 C++ implementation of above naive recursive method.

`// 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` `jobComparataor(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 inclueded ` ` ` `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, jobComparataor); ` ` ` ` ` `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; ` `}` |

*chevron_right*

*filter_none*

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 C++ implementation based on Dynamic Programming.

`// 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` `jobComparataor(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, jobComparataor); ` ` ` ` ` `// 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; ` `} ` |

*chevron_right*

*filter_none*

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

## Recommended Posts:

- Assembly Line Scheduling | DP-34
- Shortest path with exactly k edges in a directed and weighted graph
- Weighted Job Scheduling in O(n Log n) time
- Find Jobs involved in Weighted Job Scheduling
- Weighted Job Scheduling | Set 2 (Using LIS)
- Number of ways to get a given sum with n number of m-faced dices
- Total number of odd length palindrome sub-sequence around each centre
- Number of sub-sequence such that it has one consecutive element with difference less than or equal to 1
- Iterative approach to print all permutations of an Array
- Maximum sum of non-overlapping subarrays of length atmost K
- Iterative approach to print all combinations of an Array
- Longest sub-sequence that satisfies the given conditions
- Maximum subset sum such that no two elements in set have same digit in them
- Number of ways to choose elements from the array such that their average is K