20 Must Practice basic Java Programs for beginners

Write a program to swap two numbers in Java.
Ans. Two ways to do this -with third variable and without third variable.

public static void swapNumberswithtemp(int a, int b) { //using 3rd variable
int temp = a;
a = b;
b = temp;
}
public static void swapNumberswithouttemp(int a, int b) {//without using 3rd variable
b = b + a;
a = b - a;
b = b - a;
}
public static void main(String[] args) {
int a = 10;
int b = 20;
swapNumbers(a, b);
System.out.printf("%d %d", a, b);

Write a program to print all the elements of Fibonacci series.
Ans. A Fibonacci series is given by Fib(n)=Fib(n-1)+Fib(n-2)

public static int fibonacci(int number){
    if(number == 1 || number == 2){ //base case
        return 1;
    }
    return fibonacci(number-1) + fibonacci(number -2); 
}  
public static void main(String args[]) {
    int number = new Scanner(System.in).nextInt();
    for(int i=1; i<=number; i++){
        System.out.print(fibonacci2(i) +" ");
    }    
}

Check if a given number is palindrome or not.
Ans. A palindrome is a number which is when reversed give the same number. We find the reverse and check whether it is equal to the given number or not.

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    String str = in.nextLine();
    int length = str.length();         
    boolean isPalindrome = true;          
    for(int i = 0; i < length; i++)
    {
        if(str.charAt(i) != str.charAt(length-1-i)) {
            System.out.println("Snot a palindrome.");
            isPalindrome = false;
            break;
        }
    }          
    if(isPalindrome) {
        System.out.println("palindrome.");
    }
}

Write a program to find whether a number is an Armstrong number or not.
Ans. Armstrong number – Sum of the cubes of its digit is equal to the number itself. E.g - 153

public static void main(String[] args) 
{
    int num = Integer.parseInt(args[0]);
    int temp = num;
    int count = 0;
    int sum = 0;
    while(num > 0)   
    {
        num = num / 10;
        count++;
    }
    num = temp;
    while (num > 0)
    {
        int rem = num % 10;
        int val = 1;
        for(int i = 1; i <= count; i++)
        {   
            val = val * rem;
        }
        num = num / 10;
        sum = sum + val;
    }   
    if(temp == sum)
        {
            System.out.println("Armstrong number");
        }
        else
        {
            System.out.println("Not an armstrong number");
        }   
}

Find the GCD of two numbers in logn time.
Ans.

public static int gcd(int a, int b) {
while (((a > 0) && (b > 0))) {
  if ((a > b)) {
    a = (a % b);
  } else {
    b = (b % a);
  }
}
if ((a == 0)) {
  return b;
} else {
  return a;
}
}  
public static void main(String[] args) {
int a = new Scanner(System.in);
int b = new Scanner(System.in);
System.out.println(GCD.gcd(a, b));
}

Write a program to find the sum of n natural numbers
Ans.

summation of i from 1 to n=n*(n+1)/2
public static void main(String arg[]){
              int n,sum=0;
              Scanner sc=new Scanner(System.in);
                sum = (n*(n+1))/2;
               System.out.println(sum);                  
            }

Write a program to find the lcm of two numbers.
Ans.

private static int getGCD(int num1, int num2) {
while (num2 > 0) {
    int temp = num2;
    num2 = num1 % num2;
    num1 = temp;
}
return num1;
}
private static int getLCM(int num1, int num2) {
return num1 * (num2 / getGCD(num1, num2));
}

Calculate the sum of digits of a given number.
Ans.

public static void main(String args[]){
 long number=-9999,input=-9999;
    try{    
       number=Long.parseLong(System.console().readLine());
       input=number;    
    }catch(NumberFormatException nfe){
       System.exit(1);    
      }
    long remainder=0,total=0;
    while(number > 0){
        remainder=number % 10;
        total+=remainder;
        number=number/10;
     }
    System.out.println(total);
}

Write a program to reverse a string.
Ans. We use the string buffer class to easily reverse the string

public static void main(String args[]) {
    String word = "HelloWorld";
    String reverse = new StringBuffer(word).reverse().toString();
    word = "thisisstring";
    reverse = new StringBuilder(word).reverse().toString();
    System.out.printf(reverse);
}     
public static String reverse(String source){
    if(source == null || source.isEmpty()){
        return source;
    }       
    String reverse = "";
    for(int i = source.length() -1; i>=0; i--){
        reverse = reverse + source.charAt(i);
    }      
    return reverse;
}

Can you find the second highest number in given array?
Ans.

public static void main(String[] args)
{
    int arr[] = { 1,4,2,3,5};
    int largest = 0;
    int secondLargest = 0;
    for (int i = 0; i < arr.length; i++)
    {
        System.out.print(arr[i] + "\t");
    }
    for (int i = 0; i  largest)
        {
            secondLargest = largest;
            largest = arr[i];
        }
        else if (arr[i] > secondLargest)
        {
            secondLargest = arr[i];
        }
    }
    System.out.println(secondLargest);
}

Find the first repeating element In the array.
Ans.

void firstrep(int a[], int n) 
{
    Set set = new HashSet();
    int firstrepeating = -1;
    for (int i = 0; i < n; i++){            
        if (set.contains(a[i])) {
            majority = i;
        } else {
            set.add(a[i]);
        }
    }        
    System.out.println(a[firstrepeating]);
}
public static void main(String[] args) 
{
    int arr[] = {1, 2, 5, 4, 5, 5, 1, 2};
    int n = arr.length;
    firstrep(arr,n);
}

Write a program to find two elements in the array whose sum is near to zero.
Ans.

public static void main (String[] args) 
{
  int arr[] = {3, 2, -1, 9, -8, 6};
  int size = arr.length;
  int l, r, min_sum, sum, min_l_num, min_r_num;      
  min_l_num = 0;
  min_r_num = 1;
  min_sum = arr[0] + arr[1];      
  for(l = 0; l < size - 1; l++)   {
    for(r = l+1; r  Math.abs(sum))
      {
        min_sum = sum;
        min_l_num = l;
        min_r_num = r;
      }
    }
  }
  System.out.println(arr[min_l_num],arr[min_r_num]);
}

What is Singleton and write a program to describe it.
Ans. Singleton class in Java has only one instance in the whole program. We can easily implement a singleton class with thread safe property. We can see the implementation below –

public class Javasingleton
{
private static  volatile Javasingleton  singleinstance;
private Javasingleton(){}
public static   Javasingleton  getInstance()
{
    if (singleinstance ==null )
    {
        synchronized(Javasingleton.class)
        {
            if (singleinstance ==null )
            {
                singleinstance=new Javasingleton();
            }
        }
    }
    return singleinstance ;
}
}

Write a code to print all the first n prime numbers where n will be given as input.
Ans.

public static void main(String[] args) {
    int N = 200;
    int k;
    Scanner sc = new Scanner(System.in);
    k = sc.nextInt();
    int i = 0;
    for (int num = 2; num <= N; num++) {
        if (isPrimeNumber(num) && i!=k) {
            System.out.println(num);
            i++;
        }
    }
}
public static boolean isPrimeNumber(int num) {
    for (int i = 2; i 0)
      {
        RevNumber=(RevNumber*10)+(n%10);
        n=n/10;
      }
      return  (int) RevNumber;   
    }

Can you write the Bubble Sort Algorithm?
Ans. The idea is to bubble out the greater numbers to the right.

public void sort(int[] numbers) {
validateInput(numbers);
int length = numbers.length;
boolean swap = true;
while (swap) {
  swap = false;
  for (int i = 0; i < length - 1; i++) {
    for (int j = 0; j  numbers[j + 1]) {
        swap(numbers, j, j + 1);
        swap = true;
      }
    }
  }
}
}
}

Implement the merge Sort algorithm.
Ans. This is a Divide and conquer algorithm.

public static void main(String[] args) {
    int[] list = {4,5,2,1,3};
    mergeSort(list, 0, list.length - 1);
}
public static void mergeSort(int[] a, int first, int last)
  {
    if(last - first == 0) 
    {
    }
    else if (last - first == 1) 
    {
        if(a[first] > a[last])
        {
            int temp = a[first];
            a[first] = a[last];
            a[last] = temp;
        }
    }
    else    
    {
        int mid = (first + last) / 2;
        mergeSort(a, first, mid);
        mergeSort(a, mid + 1, last);
        merge(a, first, mid, last);
    }
  }
private static void merge(int[] a, int first, int mid, int last)
  {
    int[] temp = new int[last - first + 1];
    int i = first; int j = mid + 1;
    for(int k = first; k  mid || j > last)
        {
            if(i > mid && j <= last)
            {
                System.out.println("a[j]: " + a[j]);
                temp[k - first] = a[j];
                j++;
            }
            else if(i  last)
            {
                System.out.println("a[i]: " + a[i]);
                temp[k - first] = a[i];
                i++;
            }
            else
            {
                break;
            }
        }
        else
        {
            if(a[i] < a[j])
            {
                temp[k - first] = a[i];
                i++;
            }
            else
            {
                temp[k - first] = a[j];
                j++;
            }
        }
    }
    for(int count = 0; count < temp.length; count++)
    {
        a[first + count] = temp[count];
    }
  }

Write a program to get the sum of even numbers and odd number in an array.
Ans.

public static void main(String[] args) 
{
    int n, sumeven= 0, sumodd = 0;
    Scanner s = new Scanner(System.in);
    n = s.nextInt();
    int[] a = new int[n];
    for(int i = 0; i < n; i++)
    {
        a[i] = s.nextInt();
    }
    for(int i = 0; i  0)
{
bin[i++] = n%2;
   n = n/2;
}
for(int j = i-1;j >= 0;j--)
{
   System.out.print(bin[j]);
}
}

Write a program to add two binary numbers.
Ans.

public static void main(String[] args)
{     
long b1, b2;
int i = 0, carry = 0;
int[] sum = new int[10];
Scanner scanner = new Scanner(System.in);
b1 = scanner.nextLong();
b2 = scanner.nextLong();
while (b1 != 0 || b2 != 0) 
{
    sum[i++] = (int)((b1 % 10 + b2 % 10 + carry) % 2);
    carry = (int)((b1 % 10 + b2 % 10 + carry) / 2);
    b1 = b1 / 10;
    b2 = b2 / 10;
}
if (carry != 0) {
    sum[i++] = carry;
}
--i;
while (i >= 0) {
    System.out.print(sum[i--]);
}
System.out.print("\n");  
}

20 Must Practice basic Java Programs for beginners

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