Quick Prep Java Interview Archives | Page 21 of 28 | 800+ Java & Big Data Interview Q&As

Blog Archives

« 1 … 17 18 19 20 21 22 23 24 25 … 28 »

Convert Arrays to Lists and lists to array with real life example(s)

1. Why do you want to convert arrays to lists? 1. Arrays don’t play well with generics. 2. Arrays are of fixed size, and if you want to add or insert more elements then a list serves that purpose. 3. … Read more ›...

This content is for 30-Day-Full-Access, 100-Day-Full-Access, 200-Day-Full-Access, and 365-Day-Full-Access members only. Register 80+ Free Java FAQs 80+ Free Big Data FAQs

Already a member? Log in here

Convert Java Date/Time to String & String back to Date/Time

#1. Why convert a Date to String & a String back to Date

(a) Convert a String input from say a file date, so that you can perform operations like

1) Adding 5 days to the date.
2) Comparing a date like before,

…

Converting a BST Tree to a List in Java

This is the reverse of Converting an Array/List to BST in Java. It is a very common task to convert a collection type A to B as a developer. More examples to practice: Converting from A to B There are 3 ways to traverse a tree to flatten it to...

This content is for 30-Day-Full-Access, 100-Day-Full-Access, 200-Day-Full-Access, and 365-Day-Full-Access members only. Register 80+ Free Java FAQs 80+ Free Big Data FAQs

Already a member? Log in here

Converting an Array/List to BST in Java

Q. What is a BST? A. BST stands for Binary Search Tree, sometimes called ordered or sorted binary trees. This is a type of data structures that store “items” such as numbers, names etc in memory. A BST allows fast lookup, addition and removal of items, … Read more ›...

This content is for 30-Day-Full-Access, 100-Day-Full-Access, 200-Day-Full-Access, and 365-Day-Full-Access members only. Register 80+ Free Java FAQs 80+ Free Big Data FAQs

Already a member? Log in here

Converting String to Amount and Amount to String

#1. Does Java have a “Money” class? No. This will change in Java 9 with the “Money API”. JSR 354 defines a new Java API for working with Money and Currencies. #2. What are the 2 potential pitfalls in working with money? … Read more ›...

This content is for 30-Day-Full-Access, 100-Day-Full-Access, 200-Day-Full-Access, and 365-Day-Full-Access members only. Register 80+ Free Java FAQs 80+ Free Big Data FAQs

Already a member? Log in here

Event sourcing & CQRS interview Q&As

Most feasible way to handle consistency across microservices is via eventual consistency. This model doesn’t enforce distributed ACID transactions across microservices. Event sourcing is an event-centric approach to business logic design and persistence. It favours to use some mechanisms of ensuring that the system would be eventually consistent at some...

This content is for 30-Day-Full-Access, 100-Day-Full-Access, 200-Day-Full-Access, and 365-Day-Full-Access members only. Register 80+ Free Java FAQs 80+ Free Big Data FAQs

Already a member? Log in here

Fibonacci number

Q1. Can you write a function to determine the nth Fibonacci number? The Fibonacci numbers under 2000 are : 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597. Where the zeroth number being 0, first number being 1, … Read more...

This content is for 30-Day-Full-Access, 100-Day-Full-Access, 200-Day-Full-Access, and 365-Day-Full-Access members only. Register 80+ Free Java FAQs 80+ Free Big Data FAQs

Already a member? Log in here

Fibonacci number with caching and Java 8 FP

Complimenting Fibonacci number coding – iterative and recursive approach, we can improve performance by caching. If you run this

 public class RecursiveFibonacci {

  public int fibonacci(int n) {
    if (n == 0 || n == 1)
      return n;

    System.out.println("evaluating fibonacci(" + n + ")");
    return fibonacci(n - 2) + fibonacci(n - 1);
  }

  public static void main(String[] args) {
      int nThfibonacciNo = new RecursiveFibonacci().fibonacci(5); 
      System.out.println(nThfibonacciNo);
  }

}

public class RecursiveFibonacci {

public int fibonacci(int n) {

if (n == 0 || n == 1)

return n;

System.out.println("evaluating fibonacci(" + n + ")");

return fibonacci(n - 2) + fibonacci(n - 1);

}

public static void main(String[] args) {

int nThfibonacciNo = new RecursiveFibonacci().fibonacci(5);

System.out.println(nThfibonacciNo);

}

Output

 evaluating fibonacci(5)
evaluating fibonacci(3)
evaluating fibonacci(2)
evaluating fibonacci(4)
evaluating fibonacci(2)
evaluating fibonacci(3)
evaluating fibonacci(2)
5

evaluating fibonacci(5)

evaluating fibonacci(3)

evaluating fibonacci(2)

evaluating fibonacci(4)

evaluating fibonacci(2)

evaluating fibonacci(3)

evaluating fibonacci(2)

and you can see “fibonacci(3)” is repeated 2 times, “fibonacci(2)” is repeated 3 times, and so on. If you pick a larger number like 21, … Read more...

This content is for 30-Day-Full-Access, 100-Day-Full-Access, 200-Day-Full-Access, and 365-Day-Full-Access members only. Register 80+ Free Java FAQs 80+ Free Big Data FAQs

Already a member? Log in here

Find 2 numbers from an array that add up to the target

Q1. Given an array of integers, find two numbers such that they add up to a specific target number? For example, Given numbers: {2, 3, 8, 7, 5} Target number: 9 Result: 2 and 7 A1. Solution 1: Store processed numbers in a set. … Read more ›...

This content is for 30-Day-Full-Access, 100-Day-Full-Access, 200-Day-Full-Access, and 365-Day-Full-Access members only. Register 80+ Free Java FAQs 80+ Free Big Data FAQs

Already a member? Log in here

Find middle element of a LinkedList in one pass Java example

The SinglyLinkedList was created in the post “LinkedList creating from scratch in Java“. This extends “SinglyLinkedList” post. Q. How to find the middle node of a linked list in a single pass? A.

package linked.list;

public class LinkedListTest {

    public static void main(String args[]) {

        SinglyLinkedList<Integer> linkedList = new SinglyLinkedList<Integer>();
        linkedList.add(1);
        linkedList.add(2);
        linkedList.add(3);

        printMiddleNodeInsinglePass(linkedList);

    }

    public static void printMiddleNodeInsinglePass(SinglyLinkedList<Integer> linkedList) {
        Node<Integer> current = linkedList.getHead();
        int length = 0;
        Node<Integer> middle = linkedList.getHead();


        //Loop until last element is reached
        while (current.getNext() != null) {
            length++;
            if (length % 2 == 0) {
                middle = middle.getNext();
            }
            current = current.getNext();
        }

        if (length % 2 == 1) {
            middle = middle.getNext();
        }

        System.out.println("length of SinglyLinkedList: " + (length+1));
        System.out.println("middle element of SinglyLinkedList : " + middle);

    }
}

package linked.list;

public class LinkedListTest {

public static void main(String args[]) {

SinglyLinkedList<Integer> linkedList = new SinglyLinkedList<Integer>();

linkedList.add(1);

linkedList.add(2);

linkedList.add(3);

printMiddleNodeInsinglePass(linkedList);

}

public static void printMiddleNodeInsinglePass(SinglyLinkedList<Integer> linkedList) {

Node<Integer> current = linkedList.getHead();

int length = 0;

Node<Integer> middle = linkedList.getHead();

//Loop until last element is reached

while (current.getNext() != null) {

length++;

if (length % 2 == 0) {

middle = middle.getNext();

}

current = current.getNext();

}

if (length % 2 == 1) {

middle = middle.getNext();

}

System.out.println("length of SinglyLinkedList: " + (length+1));

System.out.println("middle element of SinglyLinkedList : " + middle);

}

Output:

length of SinglyLinkedList: 3
middle element of SinglyLinkedList : data=2, next=data=3, next=null

length of SinglyLinkedList: 3

middle element of SinglyLinkedList : data=2, next=data=3, next=null

This content is for 30-Day-Full-Access, 100-Day-Full-Access, 200-Day-Full-Access, and 365-Day-Full-Access members only. Register 80+ Free Java FAQs 80+ Free Big Data FAQs

Already a member? Log in here

Find pair of numbers with a given sum in Java

Q. Write a method which takes the parameters (int[ ] inputNumbers, int sum) and checks input to find the pair of integer values which totals to the sum. If found returns true, else returns false? Considerations: Should it work for negative integers? … Read more ›...

This content is for 30-Day-Full-Access, 100-Day-Full-Access, 200-Day-Full-Access, and 365-Day-Full-Access members only. Register 80+ Free Java FAQs 80+ Free Big Data FAQs

Already a member? Log in here

Find the first non repeated character in a given string input

Firstly, understand the problem, and write down any considerations & assumptions. Next write pseudocode if that helps. Considerations 1) Pre-codnition check for null or empty input. 2) Loop through the input string, and store each “character” as a key in a map with the value being the “character count”. …...

This content is for 30-Day-Full-Access, 100-Day-Full-Access, 200-Day-Full-Access, and 365-Day-Full-Access members only. Register 80+ Free Java FAQs 80+ Free Big Data FAQs

Already a member? Log in here

Finding the missing numbers Java example

Q. Can you write code to identify missing numbers in a given array of numbers?

Solution 1: Assuming that the given numbers are in order

public class MissingNumber {

    public static void main(String[] args) {
        int numbers[] = { 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 18 };
        new MissingNumber().findMissingNumbers(numbers, 0, 20);
    }

    private void findMissingNumbers(int[] input, int first, int last) {
        printMissingNumbersFront(input, first);
        printMissingNumbersInTheMiddle(input, last);
        printMissingNumbersBack(input, last);
    }

    /**
     * before 2
     * @param input
     * @param first
     */
    private void printMissingNumbersFront(int[] input, int first) {
        //0 and 1 are less than 2 as per the above input
        for (int i = first; i < input[0]; i++) {
            System.out.println(i);
        }
    }

    /**
     * after 18
     * @param input
     * @param last
     */
    private void printMissingNumbersBack(int[] input, int last) {
        // 1 + the last element value is 19. 19 to 20 are less than or equal to the last value of 20
        for (int i = 1 + input[input.length - 1]; i <= last; i++) {
            System.out.println(i);
        }
    }


    /**
     * between 2 and 18
     * @param input
     * @param last
     */
    private void printMissingNumbersInTheMiddle(int[] input, int last) {
        //for a given index i, index [i-1]+1 should be equal if present.

        for (int i = 1; i < input.length; i++) {
            //e.g. for input[1] = 3, j=1+2
            //input[2] = 4, j=1+3
            //input[3] = 6, j=4+1, Oops 5 is less than 6 so print
            //input[4] = 7, j=6+1

            //if index[i] < index[i-1] + 1, then number index[i-1] + 1 is missing
            for (int j = 1+input[i-1]; j < input[i]; j++) {
                System.out.println(j);
            }
        }
    }
}

public class MissingNumber {

public static void main(String[] args) {

int numbers[] = { 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 18 };

new MissingNumber().findMissingNumbers(numbers, 0, 20);

}

private void findMissingNumbers(int[] input, int first, int last) {

printMissingNumbersFront(input, first);

printMissingNumbersInTheMiddle(input, last);

printMissingNumbersBack(input, last);

}

/**

* before 2

* @param input

* @param first

private void printMissingNumbersFront(int[] input, int first) {

//0 and 1 are less than 2 as per the above input

for (int i = first; i < input[0]; i++) {

System.out.println(i);

}

/**

* after 18

* @param input

* @param last

private void printMissingNumbersBack(int[] input, int last) {

// 1 + the last element value is 19. 19 to 20 are less than or equal to the last value of 20

for (int i = 1 + input[input.length - 1]; i <= last; i++) {

System.out.println(i);

}

/**

* between 2 and 18

* @param input

* @param last

private void printMissingNumbersInTheMiddle(int[] input, int last) {

//for a given index i, index [i-1]+1 should be equal if present.

for (int i = 1; i < input.length; i++) {

//e.g. for input[1] = 3, j=1+2

//input[2] = 4, j=1+3

//input[3] = 6, j=4+1, Oops 5 is less than 6 so print

//input[4] = 7, j=6+1

//if index[i] < index[i-1] + 1, then number index[i-1] + 1 is missing

for (int j = 1+input[i-1]; j < input[i]; j++) {

System.out.println(j);

}

Output:

The above solution assumes that the numbers are in order (i.e.

…

Blog Archives

Convert Arrays to Lists and lists to array with real life example(s)

Convert Java Date/Time to String & String back to Date/Time

Converting a BST Tree to a List in Java

Converting an Array/List to BST in Java

Converting String to Amount and Amount to String

Event sourcing & CQRS interview Q&As

Fibonacci number

Fibonacci number with caching and Java 8 FP

Find 2 numbers from an array that add up to the target

Find middle element of a LinkedList in one pass Java example

Find pair of numbers with a given sum in Java

Find the first non repeated character in a given string input

Finding the missing numbers Java example

300+ Java Interview Questions - FAQs

10+ Companion Techs Interview Q&As

300+ Big Data Interview Questions - FAQs

16+ Java Tech Key Areas Interview Q&As

Your Career Tips to go places

Tutorials on Java & Big Data

300+ Java Interview Questions Answered

Disclaimer