Sunday, 3 September 2017

Introduction of Java

Basic Knowledge of Java programming language-

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Author: James Gosling

Vendor: Sun Micro System

Project name: Green Project

Type: open source & free software

Initial Name: OAK language

Present Name: java

Extensions : .java & .class & .jar

Initial version : jdk 1.0 (java development kit)

Present version: java 8 March 2015

Operating System : multi Operating System

Symbol: coffee cup with saucer

Objective: To develop web applications

SUN: Stanford Universally Network

Slogan/Motto: WORA(write once run anywhere)




Importance of Core Java

Importance of Core Java:-


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According to the SUN, 3 billion devices run on the Java language only.
1) Java is used to develop Desktop Applications such as MediaPlayer, Antivirus etc.
2) Java is Used to Develop Web Applications such as durgajobs.com, irctc.co.in etc.
3) Java is Used to Develop Enterprise Application such as Banking applications.
4) Java is Used to Develop Mobile Applications.
5) Java is Used to Develop Embedded System.
6) Java is Used to Develop SmartCards.
7) Java is Used to Develop Robotics.
8) Java is used to Develop Games etc.


Technologies Depends on Core Java





JAVA VERSIONS:-
------------------------------------------------------------------------------------------
Java Alpha & beta : 1995

JDK 1.0 : 1996

JDK1.1 : 1997

J2SE 1.2 : 1998

J2SE 1.3 : 2000

J2SE 1.4 : 2002

J2SE 1.5 : 2004

JAVA SE 6 : 2006

JAVA SE 7 : 2011

JAVA SE8 : 2015




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Thursday, 11 May 2017

Java Program For Find The Salary Of An Employee With Employee Grade

Problem :- Java Program For Find The Salary Of An Employee With Employee Grade (Performance).

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"write a code in java that input salary and grade of an employee and apply below conditions:
i) in case of grade 15 or above than bonus is 15%
ii) in case of grade 16 or above than bonus is 20%
iii) in case of grade 18 or above than bonus is 25%
iv) after calculating total salary deduct 13% GST in case that salary is 15000 or above. deduct 15% GST and in case that salary is 22000 or above.
v) Add 6% bonus at the end 
Calculate net salary according to above condition and display it.
"


Java Program For Find The Salary Of An Employee With Employee Grade


Logic :- So according to first three condition in problem i use if else Statement For grade of employee See Below .

if(grade==15)
{
bonus=(basic_salary*15)/100;
}
else if(grade==16 || grade==17 )
{
bonus=(basic_salary*20)/100;
}
else if(grade>=18)
{
bonus=(basic_salary*25)/100;
}

After That We Get a Basic Salary + Bonus . Now According to GST ( Goods and Service Tax ) Conditions Calculate a remaining Salary After All Calculation 6% Bonus  .

if(basic_salary>=15000 && basic_salary<22000)
{
gst=(basic_salary*13)/100;

basic_salary-=gst;

bonus1=(basic_salary*6)/100;

basic_salary+=bonus1;

}
else if(basic_salary>=22000)
{
gst=(basic_salary*15)/100;

basic_salary-=gst;

bonus1=(basic_salary*6)/100;

basic_salary+=bonus1;

}
        else
{
bonus1=(basic_salary*6)/100;

basic_salary+=bonus1;
}

At the End Print The Salary .

Example :- Lets Assume Employee Salary is 20,000 And Grade is 15 Then Calculate a Employee Salary  .

20,000 Salary and 15 Grade Bonus is 15% then Bonus =3000 .

Now Total Salary is 20,000+3000=23,000

Now Salary is Greater than 22,000 Then 15% GST ,So GST=3450 .

Now Total Salary is 23,000-3450=19550 .

At the End 6% Bonus Then Bonus = 1173

So Gross Salary Of Employee is =19500+1173=20723.

Gross Salary = 20723 .

Try This Java Program For Find The Gross Salary Of An Employee

Solution :-

import java.util.Scanner;
class employeesalary
{
 public static void main(String args[]) 
 {
float basic_salary,bonus1=0,bonus=0,gst;
int grade;

Scanner scan=new Scanner(System.in);

System.out.println("Enter Basic Salary Of Employee : ");
basic_salary=scan.nextFloat();

System.out.println("\nEnter The Grade Of Employee : ");    
grade=scan.nextInt();

if(grade==15)
{
bonus=(basic_salary*15)/100;
}
else if(grade==16 || grade==17 )
{
bonus=(basic_salary*20)/100;
}
else if(grade>=18)
{
bonus=(basic_salary*25)/100;
}

basic_salary+=bonus;

if(basic_salary>=15000 && basic_salary<22000)
{
gst=(basic_salary*13)/100;

basic_salary-=gst;

bonus1=(basic_salary*6)/100;

basic_salary+=bonus1;

}
else if(basic_salary>=22000)
{
gst=(basic_salary*15)/100;

basic_salary-=gst;

bonus1=(basic_salary*6)/100;

basic_salary+=bonus1;
}
else
{
bonus1=(basic_salary*6)/100;

basic_salary+=bonus1;
}

    System.out.println("\nGross Salary Of Employee : "+basic_salary);

  }


Output :-


Java Program For Find The Salary Of An Employee With Employee Grade

Thursday, 6 April 2017

Geeksforgeeks Solution For " Two Repeated Elements "

GeeksforGeeks Solution For Hard Domain .Below You Can Find The Solution Of  School Basic ,Easy ,Medium . Or Hackerrank Solution You Can Also Direct Submit Your Solution to Geeksforgeeks Same Problem .You Need to login then you can submit you answers 

Problem :- Two Repeated Elements

Submit Your Solution :- Click Here 

Solution :- 

#include<iostream>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--){
   int n;
   cin>>n;
   int a[n+2],b[n+2]={};

   for(int i=0;i<n+2;i++){
       cin>>a[i];
       b[a[i]]=b[a[i]]+1;
       if(b[a[i]]==2)
           cout<<a[i]<<" ";
   }
   cout<<endl;
}
return 0;
}

Output:-



Geeksforgeeks Solution For " Two Repeated Elements "

Geeksforgeeks Solution For " Sum of Middle Elements of two sorted arrays "

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Problem :- Sum of Middle Elements of two sorted arrays

Submit Your Solution :- Click Here 

Solution :- 

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main(){
int t,n,val;
cin>>t;
while(t--){

cin>>n;
int a[n],b[n];
vector<int>c(2*n);
for(int i=0;i<n;i++){
cin>>a[i];
}

for(int i=0;i<n;i++){
cin>>b[i];
}
merge(a,a+n,b,b+n,c.begin());
int s=c.size();
cout<<(c[(s/2)-1]+c[s/2])<<endl;

}
}

Output:-



Geeksforgeeks Solution For " Sum of Middle Elements of two sorted arrays "

Geeksforgeeks Solution For " Stock buy and sell "

GeeksforGeeks Solution For Hard Domain .Below You Can Find The Solution Of  School Basic ,Easy ,Medium . Or Hackerrank Solution You Can Also Direct Submit Your Solution to Geeksforgeeks Same Problem .You Need to login then you can submit you answers 

Problem :- Stock buy and sell

Submit Your Solution :- Click Here 

Solution :- 

#include<bits/stdc++.h>
using namespace std;
int main()
 {
int t;
    cin>>t;
    while(t--)
    {
        int n;
   cin>>n;
   int a[n];
   for(int i=0;i<n;i++)
       cin>>a[i];
   int buy=0,sell=0;
   bool key=false;
   for(int i=1;i<n;i++)
   {
       if(a[i]>a[i-1])
           sell++;
       else
       {
           if(buy!=sell)
           {
               cout<<"("<<buy<<" "<<sell<<") ";
               key=true;
           }
           buy=sell=i;
       }
   }
   if(buy!=sell)
   {
       cout<<"("<<buy<<" "<<sell<<") ";
        key=true;
   }
   if(!key)
        cout<<"No Profit";
   cout<<endl;       
    }
return 0;
}

Output:-



Geeksforgeeks Solution For " Stock buy and sell "

Geeksforgeeks Solution For " Reverse Coding "

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Problem :- Reverse Coding

Submit Your Solution :- Click Here 

Solution :- 

#include <iostream>
using namespace std;

int main() {
int t;
cin>>t;
while(t--){
   int n,m;
   cin>>n>>m;
   int c=(n*(n+1))/2;
   if(c==m)
       cout<<1<<endl;
        else
            cout<<0<<endl;
}
return 0;
}

Output:-



Geeksforgeeks Solution For " Reverse Coding "

Geeksforgeeks Solution For " Move all negative elements to end "

GeeksforGeeks Solution For Hard Domain .Below You Can Find The Solution Of  School Basic ,Easy ,Medium . Or Hackerrank Solution You Can Also Direct Submit Your Solution to Geeksforgeeks Same Problem .You Need to login then you can submit you answers 

Problem :- Move all negative elements to end

Submit Your Solution :- Click Here 

Solution :- 

#include <iostream>
#include <vector>
using namespace std;

int main() {
int t;
    cin>>t;
    while(t--){
        int no,p=0,n=0;
        cin>>no;
        vector<int> po,ng;
        int a[no];
        for(int i=0;i<no;i++){
            cin>>a[i];
            if(a[i]>=0){
                p++;
                po.push_back(a[i]);
            }    
            else{
                n++;
                ng.push_back(a[i]);
            }    
        }
        for(int i=0;i<p;i++)
            a[i]=po[i];
        for(int i=0;i<n;i++)    
            a[p+i]=ng[i];
        for(int i=0;i<no;i++)    
            cout<<a[i]<<" ";
        cout<<endl;    
    }
return 0;
}

Output:-



Geeksforgeeks Solution For " Move all negative elements to end "

Geeksforgeeks Solution For " Maximum repeating number "

GeeksforGeeks Solution For Hard Domain .Below You Can Find The Solution Of  School Basic ,Easy ,Medium . Or Hackerrank Solution You Can Also Direct Submit Your Solution to Geeksforgeeks Same Problem .You Need to login then you can submit you answers 

Problem :- Maximum repeating number

Submit Your Solution :- Click Here 

Solution :- 

#include <bits/stdc++.h>
using namespace std;

int main() {
int t;
cin>>t;
while(t--){
   int n, k;
   cin>>n>>k;
   int a[n], b[n]={};
   for(int i=0; i<n; i++){
       cin>>a[i];
       b[a[i]]=b[a[i]]+1;
   }
   int index=0,max_v=b[0];
           
            for(int i=0;i<k;i++)
                if(b[i]>max_v){
                    max_v=b[i];
                    index=i;
                }
            cout<<index<<endl;
}
return 0;
}

Output:-



Geeksforgeeks Solution For " Maximum repeating number "

Geeksforgeeks Solution For " Maximum no of 1's row "

GeeksforGeeks Solution For Hard Domain .Below You Can Find The Solution Of  School Basic ,Easy ,Medium . Or Hackerrank Solution You Can Also Direct Submit Your Solution to Geeksforgeeks Same Problem .You Need to login then you can submit you answers 

Problem :- Maximum no of 1's row

Submit Your Solution :- Click Here 

Solution :- 

#include <iostream>
using namespace std;

int main() {
int t;
cin>>t;
while(t--){
   int r,c,i,j,row_n,prev_c=0,cur_c;
   cin>>r>>c;
   bool a[r][c];
   for(i=0; i<r; i++)
       for(j=0; j<c; j++)
           cin>>a[i][j];
   
   for(i=0; i<r; i++){
       cur_c=0;
       for(j=0; j<c; j++){
          if(a[i][j]==1)
           cur_c++;
       }   
       if(cur_c>prev_c){
            row_n=i;
       prev_c=cur_c;
       }
               
   }     
   cout<<row_n<<endl;
}
return 0;
}

Output:-



Geeksforgeeks Solution For " Maximum no of 1's row "