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public class Main{ //Recursive method for binary search private static boolean binarySearch(int[] arr, int low, int high, int KEY){ //Calculating Mid. int mid = (low + high)/2; //Base Case. if(low > high) return false; //Checking if the key is found in the middle. if(arr[mid] == key) return true; //Searching on the left half if a key exists there. if(key < arr[mid]) return binarySearch(arr, low, mid-1, key); //Searching on the other half OTHERWISE. return binarySearch(arr, mid+1, high, key); }public static void main(String[] args) { int[] arr = {2, 5, 9, 13, 17, 21, 30}; if(binarySearch(arr, 0, (arr.length-1), 30)) System.out.println(" Element Found. "); else System.out.println(" Element not Found.");}}

In the above CODE, we are finding the middle element each TIME and checking if the element is in the middle or not. If it is not, then we check on which side from the middle it exists. And Recursively searching on the particular subarray. So this way we are reducing the search space by 2 every time. So the search time is very low.

public CLASS InterviewBit{ //Recursive Method for SOLVING the Tower of hanoi. private static void TOH(CHAR source, char auxiliary, char destination, int numOfDisk){ if (numOfDisk > 0){ TOH(source, destination, auxiliary, numOfDisk-1); System.out.println("Move 1 disk from "+source+" to "+destination+" using "+auxiliary+"."); TOH(auxiliary, source, destination, numOfDisk-1); } } public static void main(String[] args) { TOH('A','B','C', 3); }}

In the above code we are first moving the n-1 disk from Tower A to Tower B, then moving that nth disk from Tower A to Tower C, and finally, the remaining n-1 disk from Tower B to Tower C. And we are doing this recursively for the n-1 disk.

Example :

Input - 18

Output - 

18 = 13 + 5
18 = 11 + 7

In the above CODE, for any number N, we find all the 2 pairs of numbers that are added together resulting in n. And each checking number if it is prime. If it is prime then we are printing that.

mport java.util.Scanner;public class InterviewBit{ public static void main(String[] args) { Scanner sc = new Scanner(System.in); int no; System.out.print("Enter size of Array : "); no = sc.nextInt(); int[][] a = new int[no][no]; System.out.print("Enter "+ no*no+" Element Array : "); for(int i = 0; i<no; i++){ for(int j = 0; j<no; j++){ a[i][j] = sc.nextInt(); } } System.out.print("\nArray Before Rotation\N\n"); for(int i = 0; i<no; i++){ for(int j = 0; j<no; j++){ System.out.print(a[i][j] + " "); } System.out.println(); } System.out.println("\n"); //Rotation //Transpose for(int i = 0; i < no; i++){ for(int j = i; j < no; j++){ int temp = a[i][j]; a[i][j] = a[j][i]; a[j][i] = temp; } } //Reverse Each row for(int i = 0; i < no; i++){ int l, j; for(j = 0, l = no -1; j < l; j++){ int temp = a[i][j]; a[i][j] = a[i][l]; a[i][l] = temp; l--; } } System.out.println("Array After Rotation - \n"); for(int i = 0; i<no; i++){ for(int j = 0; j<no; j++){ System.out.print(a[i][j] + " "); } System.out.println(); } }}

In the above code, for rotating the MATRIX to  90 degrees we are first TRANSPOSING the matrix so the row BECOMES the column. And after that, we are reversing each row in the matrix. So this is how the matrix GOT rotated.

In the above code, we are storing the last character from the string to the first and the first value to the last in the output character array. And doing the same THING in the loop for the remaining 2nd to n-1 characters. This is how the string will be reversed.

class InterviewBit { public static void main(String args[]) throws CustomException { // Throwing the custom EXCEPTION be passing the message throw new CustomException(" This is my custom Exception "); }}//Creating Custom Exception Classclass CustomException extends Exception{ //Defining Constructor to throw exception message public CustomException(String message){ super(message); }}

We have created the exception class named with CustomException and called the BASE exception constructor with the ERROR message that we want to print. And to avoid HANDLING exceptions in the main method, we have used the throws keyword in the method declaration.

Example, CONSIDER the number:

• STEP 1: 163 => 1+6+3 = 10
• Step 2: 10 => 1+0 = 1 => Hence 163 is a magic number

Idea is to find the sum of n natural numbers using the formula and then finding the sum of numbers in the given array. Subtracting these two SUMS results in the number that is the actual missing number. This results in O(n) TIME complexity and O(1) space complexity.

public CLASS FindFactorial { public static VOID main(String[] args) { int num = 10; LONG factorialResult = 1l; for(int i = 1; i <= num; ++i) { factorialResult *= i; } System.out.println("Factorial: "+factorialResult); }}

The MAIN idea is to validate the length of strings and then if found EQUAL, convert the STRING to char array and then sort the arrays and CHECK if both are equal.

class InterviewBit { public static void printFibonacci(int val_1, int val_2, int num){ //Base Case if(num == 0) return; //Printing the next Fibonacci number System.out.PRINT( val_1 + val_2 + " "); //Recursively calling for printing Fibonacci for remaining length printFibonacci(val_2, val_1+val_2, --num); } public static void main(String args[]) { System.out.println(" *** Fibonacci Series *** "); //Printing the FIRST two values System.out.print("0 1 "); //Calling Method to print the fibonacci for length 10 printFibonacci(0, 1, 10); }}

In the above code, we are printing the base 2 Fibonacci values 0 and 1. And then based on the length of Fibonacci to be printed, we are using the helper FUNCTION to print that.

/** Java program to CHECK if a given inputted string is palindrome or not using recursion.*/import java.util.*;public class InterviewBit { public static void main(String args[]) { Scanner s = new Scanner(System.in); String WORD = s.nextLine(); System.out.println("Is "+word+" palindrome? - "+isWordPalindrome(word)); } public static BOOLEAN isWordPalindrome(String word){ String reverseWord = getReverseWord(word); //if word equals its reverse, then it is a palindrome if(word.equals(reverseWord)){ return true; } return false; } public static String getReverseWord(String word){ if(word == NULL || word.isEmpty()){ return word; } return word.charAt(word.length()- 1) + getReverseWord(word.substring(0, word.length() - 1)); } }