Question

There are N employees already working for a company and M new candidates eligible for the work. At the end of the financial year, each of the N employees demand for an increment. You are provided with the data of current_salary and the salary he/she expects. Also, there are M candidates we can recruit. We have the data of salary demanded by each eligible candidate. The company can spend maximum of X units of money. Now you have to decide whether we should give increment to the existing employee or hire a new employee. Our aim is to maximise the staff, considering that existing staff and new applicants are at same level of knowledge and experience. Input Format: First line: 3 integers N, M, X Each of the next N lines contains two integers di and bị, where ai is the current salary of ith employee and bi is the salary he/she expects. Next line contains M integers Ci, corresponding to the salary demanded by ith candidate. Output Format: Output a single integer denoting total number of employees working for the company after this process.
Please code in java

Answer 1

Java Code:

import java.util.*;
public class Main
{
   public static void main(String[] args) {
   Scanner sc = new Scanner(System.in); //scanner object
       int i,n,m,x;   
       n=sc.nextInt(); //inputing the values of n,m,x
       m=sc.nextInt();
       x=sc.nextInt();
       int arr[][]=new int[n][x]; //two dimensional array that takes
       int inc[]=new int[n]; //previous salary and demanded salary
       for(i=0;i<n;i++)
       {
       arr[i][0]=sc.nextInt(); //inputing the values
       arr[i][1]=sc.nextInt();
       inc[i]=arr[i][1]-arr[i][0]; //inc[i] is increment in salary
       }
       int ne[]=new int[m]; //array to store values of new employees
       for(i=0;i<m;i++)
       {
       ne[i]=sc.nextInt(); //ne[i] is salary of new employees
       }
       int all[]=new int[m+n]; //all[] array to store all values of increment of old employees
       for(i=0;i<n;i++) //and salary for new employees
       {
       all[i]=inc[i];
       }
       for(;i<m+n;i++)
       {
       all[i]=ne[i-n];
       }
       Arrays.sort(all); //sorting the array to know that which operation is better
       int res=0,c=0; // increment the salary or hire new employees
       for(i=0;i<m+n;i++)
       {
       if(all[i]+res<=x)
       {
       res+=all[i];
       c++; //count number of exployees we can maximize
       }
       }
       System.out.println(c); //printing the result
   }
}

Screenshots:

import java.util.*; public class Main public static void main(String[] args) { Scanner sc = new Scanner(System.in); //scanner object int i,n,n,x; n=sc.nextInt(); //inputing the values of n,m,xi m=sc.nextInt(); x=sc.nextInt(); int arr[][]=new int[n][x]; //two dimensional array that takes int inc[]=new int[n]; //previous salary and demanded salary for(i=0;i<n;i++) arr[i][0]=sc.nextInt(); //inputing the values arr[i][1]=sc.nextInt(); inc[i]=arr[i][1] arr[i][0]; //inc[i] is increment in salary //array to store values of new employees int ne[]=new int[m]; for(i=0;i<m;i++) ne[i]=sc.nextInt(); //ne[i] is salary of new employees int all[]=new int[m+n]; for(i=0;i<n;i++) //all[] array to store all values of increment of old employees //and salary for new employees all[i]=inc[i];

for(;i<m<n;i++) all[i]=ne[i-n]; Arrays.sort(all); int res=0,c=0; for(i=0;i<m+n;i++) 1/sorting the array to know that which operation is better // increment the salary or hire new employees if(all[i]+res<=x) res+=all[i]; C++; //count number of exployees we can maximize System.out.println(c); //printing the result Activate Win

Output:

53 20 10 15 78 B 10 15 17 200 300 10 27

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