Weighted Job Scheduling

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. 

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. 

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