Create Job Scheduling_Greedy approach

Job Scheduling_Greedy approach

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+#include <iostream>
+#include <vector>
+#include <algorithm>
+using namespace std;
+
+// Data structure to store a Job
+struct Job {
+	int start, finish, profit;
+};
+
+// Function to perform binary search on the given jobs which are
+// sorted by finish time. The function returns index of the last job
+// which doesn't conflict with the given job i.e. whose finish time
+// is less than or equal to the start time of the given job.
+int findLastNonConflictingJob(vector<Job> jobs, int n)
+{
+	// search space
+	int low = 0;
+	int high = n;
+
+	// iterate till search space is exhausted
+	while (low <= high)
+	{
+		int mid = (low + high) / 2;
+		if (jobs[mid].finish <= jobs[n].start)
+		{
+			if (jobs[mid + 1].finish <= jobs[n].start)
+				low = mid + 1;
+			else
+				return mid;
+		}
+		else {
+			high = mid - 1;
+		}
+	}
+
+	// return negative index if no non-conflicting job is found
+	return -1;
+}
+
+// Function to print the non-overlapping jobs involved in maximum profit
+// using Dynamic Programming
+void findMaxProfitJobs(vector<Job> &jobs)
+{
+	// sort jobs in increasing order of their finish times
+	sort(jobs.begin(),
+		jobs.end(),
+		[](Job &x, Job &y) {
+			return x.finish < y.finish;
+		});
+
+	// get number of jobs
+	int n = jobs.size();
+
+	// maxProfit[i] stores the maximum profit possible for the first i jobs and
+	// tasks[i] stores the index of jobs involved in the maximum profit
+	int maxProfit[n];
+	vector<int> tasks[n];
+
+	// initialize maxProfit[0] and tasks[0] with the first job
+	maxProfit[0] = jobs[0].profit;
+	tasks[0].push_back(0);
+
+	// fill tasks[] and maxProfit[] in bottom-up manner
+	for (int i = 1; i < n; i++)
+	{
+		// find the index of last non-conflicting job with current job
+		int index = findLastNonConflictingJob(jobs, i);
+
+		// include the current job with its non-conflicting jobs
+		int currentProfit = jobs[i].profit;
+		if (index != -1) {
+			currentProfit += maxProfit[index];
+		}
+
+		// if including the current job leads to the maximum profit so far
+		if (maxProfit[i-1] < currentProfit)
+		{
+			maxProfit[i] = currentProfit;
+
+			if (index != -1) {
+				tasks[i] = tasks[index];
+			}
+			tasks[i].push_back(i);
+		}
+
+		// excluding the current job leads to the maximum profit so far
+		else {
+			tasks[i] = tasks[i-1];
+			maxProfit[i] = maxProfit[i-1];
+		}
+	}
+
+	// maxProfit[n-1] stores the maximum profit
+	cout << "The maximum profit is " << maxProfit[n-1] << endl;
+
+	// tasks[n-1] stores the index of jobs involved in the maximum profit
+	cout << "The jobs involved in the maximum profit are: ";
+	for (int i: tasks[n-1])
+	{
+		cout << "(" << jobs[i].start << ", " << jobs[i].finish << ", "
+			 << jobs[i].profit << ") ";
+	}
+}
+
+int main()
+{
+	vector<Job> jobs {
+		{ 0, 6, 60 },
+		{ 1, 4, 30 },
+		{ 3, 5, 10 },
+		{ 5, 7, 30 },
+		{ 5, 9, 50 },
+		{ 7, 8, 10 }
+	};
+
+	findMaxProfitJobs(jobs);
+
+	return 0;
+}