Solve the Weighted Job Scheduling problem using dynamic programming.?

The Weighted Job Scheduling problem is a common problem in computer science where we have N jobs, each with a start and end time, and a weight (profit) associated with completing the job. The objective is to find the maximum profit subset of mutually compatible jobs, where the jobs are compatible if they do not overlap in time.

We can solve this problem using dynamic programming by defining a subproblem and building a solution for the larger problem using the solutions to the subproblems. Let’s write an algorithm and code in Java to solve this problem.

Algorithm:

1. Sort the jobs by their end times in non-decreasing order so that we can easily find the compatible jobs.

2. For each i (0<=i<N), let DP[i] be the maximum profit we can obtain by scheduling the jobs from 0 to i.

3. DP[0] = jobs[0].weight

4. For i from 1 to N-1, find the largest index j (0<=j<i) such that jobs[j] does not overlap with jobs[i] and set DP[i] = max(j, DP[j] + jobs[i].weight).

5. Return DP[N-1] as the maximum profit.

Code:

Here, we assume that we have a class "Job" with start and end times and weight attributes.

public int weightedJobScheduling(Job[] jobs) {
    // Step 1: sort the jobs by their end times
    Arrays.sort(jobs, (a, b) -> a.end - b.end);
    
    // Step 2: define DP array
    int N = jobs.length;
    int[] DP = new int[N];
    
    // Step 3: initialize DP[0]
    DP[0] = jobs[0].weight;
    
    // Step 4: fill DP array
    for (int i = 1; i < N; i++) {
        DP[i] = jobs[i].weight;
        for (int j = 0; j < i; j++) {
            if (jobs[j].end <= jobs[i].start) {
                DP[i] = Math.max(DP[i], DP[j] + jobs[i].weight);
            }
        }
    }
    
    // Step 5: return DP[N-1]
    return DP[N-1];
}

Example:

Suppose we have the following jobs:

| Job ID | Start Time | End Time | Weight |
|--------|------------|----------|--------|
|   J1   |      1     |     3    |   5    |
|   J2   |      2     |     5    |   6    |
|   J3   |      4     |     6    |   5    |
|   J4   |      6     |     7    |   4    |
|   J5   |      5     |     8    |   11   |
|   J6   |      7     |     9    |   2    |

Applying the algorithm in the previous code, we obtain DP = [5, 6, 6, 10, 16, 16]. The maximum profit is 16, which is obtained by selecting jobs J1, J2, and J5.