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.
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 the longer schedule 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 myfunction(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, myfunction);       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; }

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*2n). 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 myfunction(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, myfunction);       // 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; }

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.