added a Job Sequencing Problem

Sorting/mergeSort.java

@@ -1,72 +1,86 @@
+class Sort 
+{
+    void merge(int arr[], int left, int middle, int right)
+    {
+        int low = middle - left + 1;                   
+        int high = right - middle;                      
+
+        int L[] = new int[low];                             
+        int R[] = new int[high];
 
-public class MergeSort {
-
-    public static void main(String args[]) {	
-		int arr[] = {23, 12, 85, 39, 98, 7, 16, 97, 54};
-
-		System.out.println("Given Array");
-		printArray(arr);
-
-		MergeSort ob = new MergeSort();
-		ob.sort(arr, 0, arr.length - 1);
-
-		System.out.println("\nSorted array");
-		printArray(arr);
-	}
-
-    void merge(int arr[], int low, int mid, int up) {	
-		int n1 = mid - low + 1;
-		int n2 = up - mid;
-
-		int Left[] = new int[n1];
-		int Right[] = new int[n2];
-
-		for (int i = 0; i < n1; ++i)
-			Left[i] = arr[low + i];
-		for (int j = 0; j < n2; ++j)
-			Right[j] = arr[mid + 1 + j];
-
-		int i = 0, j = 0;
-
-		int k = low;
-		while (i < n1 && j < n2) {
-			if (Left[i] <= Right[j]) {
-				arr[k] = Left[i];
-				i++;
-			}
-			else {
-				arr[k] = Right[j];
-				j++;
-			}
-			k++;
-		}
-		while (i < n1) {
-			arr[k] = Left[i];
-			i++;
-			k++;
-		}
-		while (j < n2) {
-			arr[k] = Right[j];
-			j++;
-			k++;
-		}
-	}
+        int i = 0, j = 0;
+
+        for (i = 0; i < low; i++)                               
+        {
+            L[i] = arr[left + i];
+        }
+        for (j = 0; j < high; j++)                             
+        {
+            R[j] = arr[middle + 1 + j];
+        }
+
+
+        int k = left;                                          
+        i = 0;                                           
+        j = 0;
 
-	void sort(int arr[], int low, int up) {	
-		if (low < up) {
-			int mid =low + (up-low)/2;
+        while (i < low && j < high)                     
+        {
+            if (L[i] <= R[j]) 
+            {
+                arr[k] = L[i];
+                i++;
+            }
+            else 
+            {
+                arr[k] = R[j];
+                j++;
+            }
+            k++;
+        }
+
+        while (i < low)                          
+        {
+            arr[k] = L[i];
+            i++;
+            k++;
+        }
+
+        while (j < high)                       
+        {
+            arr[k] = R[j];
+            j++;
+            k++;
+        }
+    }
+
 
-			sort(arr, low, mid);
-			sort(arr, mid + 1, up);
+    void mergeSort(int arr[], int left, int right)      
+    {
+        int middle;
+        if (left < right) {                             
+            middle = (left + right) / 2;
+
+            mergeSort(arr, left, middle);                    
+            mergeSort(arr, middle + 1, right);              
+
+            merge(arr, left, middle, right);               
+        }
+    }
+
+    void display(int arr[])                
+    {  
+        for (int i=0; i<arr.length; ++i) 
+        {
+            System.out.print(arr[i]+" ");
+        } 
+    } 
 
-			merge(arr, low, mid, up);
-		}
-	}
-
-	static void printArray(int arr[]) {	
-		int n = arr.length;
-		for (int i = 0; i < n; ++i)
-			System.out.print(arr[i] + " ");
-		System.out.println();
-	}	
+    public static void main(String args[])
+    {
+        int arr[] = { 9, 3, 1, 5, 13, 12 };
+        Sort ob = new Sort();
+        ob.mergeSort(arr, 0, arr.length - 1);
+        ob.display(arr);
+    }
 }

jobSequencingProbem.cpp

@@ -0,0 +1,71 @@
+//  You are given a set of N jobs where each job comes with a deadline and profit. 
+//  The profit can only be earned upon completing the job within its deadline. 
+//  Find the number of jobs done and the maximum profit that can be obtained. 
+//  Each job takes a single unit of time and only one job can be performed at a time.
+
+// Basic Outline of the approach:-
+
+// Sort the jobs in descending order of profit. 
+// If the maximum deadline is x, make an array of size x .Each array index is set to -1 initially as no jobs have been performed yet.
+// For every job check if it can be performed on its last day.
+// If possible mark that index with the job id and add the profit to our answer. 
+// If not possible, loop through the previous indexes until an empty slot is found.
+
+// Time Complexity: O(N log N) + O(N*M).
+// Space Complexity: O(M) 
+
+#include<bits/stdc++.h>
+
+using namespace std;
+// A structure to represent a job 
+struct Job {
+   int id; // Job Id 
+   int dead; // Deadline of job 
+   int profit; // Profit if job is over before or on deadline 
+};
+class Solution {
+   public:
+      bool static comparison(Job a, Job b) {
+         return (a.profit > b.profit);
+      }
+   //Function to find the maximum profit and the number of jobs done
+   pair < int, int > JobScheduling(Job arr[], int n) {
+
+      sort(arr, arr + n, comparison);
+      int maxi = arr[0].dead;
+      for (int i = 1; i < n; i++) {
+         maxi = max(maxi, arr[i].dead);
+      }
+
+      int slot[maxi + 1];
+
+      for (int i = 0; i <= maxi; i++)
+         slot[i] = -1;
+
+      int countJobs = 0, jobProfit = 0;
+
+      for (int i = 0; i < n; i++) {
+         for (int j = arr[i].dead; j > 0; j--) {
+            if (slot[j] == -1) {
+               slot[j] = i;
+               countJobs++;
+               jobProfit += arr[i].profit;
+               break;
+            }
+         }
+      }
+
+      return make_pair(countJobs, jobProfit);
+   }
+};
+int main() {
+   int n = 4;
+   Job arr[n] = {{1,4,20},{2,1,10},{3,2,40},{4,2,30}};
+
+   Solution ob;
+   //function call
+   pair < int, int > ans = ob.JobScheduling(arr, n);
+   cout << ans.first << " " << ans.second << endl;
+
+   return 0;
+}