Maximum Profit in Job Scheduling

class Solution {
public:
    int calculateMaximumProfit(vector<vector<int>>& tasks) {
        int taskCount = tasks.size(), bestProfit = 0;
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> minpq;

        for (int i = 0; i < taskCount; i++) {
            int start = tasks[i][0], finish = tasks[i][1], currentProfit = tasks[i][2];
            while (!minpq.empty() && start >= minpq.top()[0]) {
                bestProfit = max(bestProfit, minpq.top()[1]);
                minpq.pop();
            }
            minpq.push({finish, currentProfit + bestProfit});
        }

        while (!minpq.empty()) {
            bestProfit = max(bestProfit, minpq.top()[1]);
            minpq.pop();
        }

        return bestProfit;
    }

    int maxProfitScheduler(vector<int>& startTimes, vector<int>& endTimes, vector<int>& gains) {
        vector<vector<int>> arrangedTasks;
        for (size_t idx = 0; idx < gains.size(); idx++) {
            arrangedTasks.push_back({startTimes[idx], endTimes[idx], gains[idx]});
        }
        
        sort(arrangedTasks.begin(), arrangedTasks.end());
        return calculateMaximumProfit(arrangedTasks);
    }
};

Here's a summary of the solution for the problem that focuses on maximizing profit by scheduling jobs:

The provided C++ code defines a class Solution that includes two main functions for solving the job scheduling problem to achieve maximum profit:

1. calculateMaximumProfit:

   • Accepts a vector of vectors (tasks), with each inner vector representing a job characterized by a start time, end time, and the profit associated with completing the job.
   • Utilizes a priority queue to maintain the current set of tasks being considered, ensuring that the queue operates in a minimum fashion based on job end times.
   • Iterates over each job, updating the best profit that can be obtained by either taking the current job into consideration or skipping it based on previously considered jobs.

2. maxProfitScheduler:

   • Accepts separate vectors for start times, end times, and profits of various jobs.
   • Constructs a vector of tasks where each task is a vector consisting of start time, end time, and profit.
   • Sorts these tasks based on start times to prepare them for processing.
   • Calls calculateMaximumProfit with the sorted tasks to determine the maximum profit.

Overall, the implementation strategically combines dynamic programming with a greedy approach by leveraging a priority queue. This allows the function to efficiently determine the maximum profit that can be attained considering job overlaps and scheduling constraints. The use of sorting and a priority queue ensures that at each step, the decision-making process about which job to select next is optimized for the best possible outcome in terms of profit.

class Solution {
    class JobComparator implements Comparator<ArrayList<Integer>> {
        public int compare(ArrayList<Integer> job1, ArrayList<Integer> job2) {
            return job1.get(0) - job2.get(0);
        }
    }

    private int calculateMaximumProfit(List<List<Integer>> jobsList) {
        int jobsCount = jobsList.size(), greatestProfit = 0;
        PriorityQueue<ArrayList<Integer>> priorityQueue = new PriorityQueue<>(new JobComparator());

        for (int i = 0; i < jobsCount; i++) {
            int start = jobsList.get(i).get(0), finish = jobsList.get(i).get(1), jobProfit = jobsList.get(i).get(2);

            while (!priorityQueue.isEmpty() && start >= priorityQueue.peek().get(0)) {
                greatestProfit = Math.max(greatestProfit, priorityQueue.peek().get(1));
                priorityQueue.poll();
            }

            ArrayList<Integer> newJob = new ArrayList<>();
            newJob.add(finish);
            newJob.add(jobProfit + greatestProfit);
            priorityQueue.add(newJob);
        }

        while (!priorityQueue.isEmpty()) {
            greatestProfit = Math.max(greatestProfit, priorityQueue.peek().get(1));
            priorityQueue.poll();
        }

        return greatestProfit;
    }

    public int jobScheduling(int[] startTime, int[] endTime, int[] jobProfit) {
        List<List<Integer>> jobsOverview = new ArrayList<>();
        int numJobs = jobProfit.length;
        for (int i = 0; i < numJobs; i++) {
            ArrayList<Integer> jobDetails = new ArrayList<>();
            jobDetails.add(startTime[i]);
            jobDetails.add(endTime[i]);
            jobDetails.add(jobProfit[i]);
            jobsOverview.add(jobDetails);
        }

        jobsOverview.sort(Comparator.comparingInt(job -> job.get(0)));
        return calculateMaximumProfit(jobsOverview);
    }
}

This Java solution to the Maximum Profit in Job Scheduling problem uses dynamic programming with a greedy approach. Start by defining custom classes and helper methods to handle the job-related data and computations effectively. Follow this step-by-step approach embedded in the code provided:

1. Create a JobComparator class that implements Comparator<ArrayList<Integer>>. This class defines a comparison logic based on job start times to prioritize jobs with earlier start times.
2. Define the calculateMaximumProfit method, taking a list of jobs and calculating the maximum possible profit:
   • Initialize a priority queue to handle jobs dynamically, ensuring that the job with the earliest end time remains at the front.
   • Traverse through each job, cleaning up the priority queue by removing completed jobs before the current job's start time and updating the current greatest profit.
   • For each job, calculate the accumulated profit if this job is taken, then push this job and its cumulative profit to the queue.
   • Iterate through the remaining elements in the queue to confirm the maximum profit obtained.
3. The jobScheduling method sets up the problem:
   • It uses input arrays startTime, endTime, and jobProfit to create a structured list of jobs where each job is represented as a list containing start time, end time, and profit.
   • Sort the list of jobs based on start times using a lambda-based comparator.
   • Call calculateMaximumProfit with this sorted list to find and return the maximum profit.

The priority queue handles dynamic job processing efficiently, allowing the algorithm to consistently update the best possible profit scenario based on job overlaps and sequential job scheduling. Combining sorting with a clever use of priority data structures optimizes the solution for complex scenarios involving multiple overlapping jobs.