Spring and core java mock interview learnings

   Authority to spring to create objects

  @Component  -- for DAO layer works fine but readability/visibility wise its bad

  @Service --  Service layer

  @Controller -- Business Logic

  @Repository -- we use this annotation to DAO layer -- Benifits this helps to handle persistance level exceptions that spring provides

  DataAccessException is the root  class which spring provides for handling exceptions

@PostConstruct

  Called first during initialisation(Creating data base connections)

  @Predestroy  

  called B4 the container is destroyed and bean is out of container(Close all  the DB connections/clean up work)

  

Rest client

  ---------------------

  Rest template --- Client to consume the data (make a call to get , put,post)

  getForEntity and getForObject -- methods belong to resttemplate

  getForObject -- returns object

  getForEntity         -- returns header,body , what ever we want

Exchange method - same purpose belongs to rest template

While wrting an API , what are the advantages of returning entity??

We can send http status and body and headers

For delete API, what kinld of status code you will return

204 /200

Time out when ever we are going for rest template?

Basically when connecting to server connections are made with threads.

Thread is blocked if the connection is not establised and then another user tries and another thread is blocked .

So just to avoid that time out is needed so that thread is released

Can we have many threads inside the server?

yes

Server time out -- connection not established

Read time out --  connection established but not able to read some data

Versioning in rest

---------------------------

Adding a new functionality to the API and user should use the new one

Can be defined using request parameters and can be mentioned in headers

URi also it can be defined

How authentication in rest web services and how we make rest web servives secure?

Security configuration class which will be extending securityconfigurationAdapter class

Hitting a request goes to authentication filter which will give us authentication object and this object is paased to authentication builder which will provide proper authentication provider which will validate this object.

Some of the authentication providers are DAO Authentication Provider, we can define even our own

Authentication provider internally passes back to builder which will pass to security holder So next time when user logs in object will be checked by the security holder and user will be allowed to sign in

Header we pass the tokens , like jwt tokens which will be used for authentication and server will validate it.These are stateless i.e server wont store any inf and client will alwyas send relavant information via header.

Bearer is written in jwt token??Means syou are the owner .JWT used for authorisation .Once we give uname and pwd we get authorisation token which weill again be sending to the server which will in turn verify the token

  

In basic authentication where we give username and pwd where we can use base64 encoded

Content Negotiation?

What kinld of content API should accept/produce.It can be achieved using MediaType

Cross Cutting Concerns in spring?And do you implement that?

Belongs to spring AOP used for logging purpose .We want to log after the particular method or so..

Monitoring , secutity.after one method we have to perform security then it can be done

In micro services suppose we want to implement logging,tracing??Where will be implementing these?

At gateway level we ill be doing

Custom validators in spring can be created by implementing validator class

There is a method over ride and write the logic

Core Java

------------------

We have define ArrayList as Size 2 and then add 3 elements ..Will third elemt be added?

Yes it will be added as the arraylist dynamicallu grows in size

1 million string objects to array list added with out specifying size and in a for loop you are adding then it takes total og 613 ms to complete else if you specify size then it will run in 193 ms.

Internally it regrows and calls Arrays.Copyof which will create a new Array and then copy all the elements, so suppose if we have lot of elements and if its size is specified then it allocates so much and then no copy of array or new array creation is needed

Fail fast and fail safe collections

----------------------------------------

Iterators 

fail fast used by arraylist/ linked list --During itration if we try to add /modify it gives concurrent modification exception 

Fail safe : Thread safe ..CopyOnWriteArrayList , Concurrent hash map ...Modification during iteration it works perfectly fine

Fail safe is going to consume more memory , modification operations will be perfomred on a copy of collections not on the orinal collections

Arrays.asList will give us an unmodifyable list , which we cannoot alter 

SingleTon Design pattern

--------------------------

Object for the class can be created only one(For the entire run time)

Scope Single ton in spring means one bean per container(one per application context)

Print Zero,Even,Odd numbers in this order(0,1,0,2,0,3,0,4...).Threads started at the same time

import java.util.concurrent.Semaphore;

import java.util.function.IntConsumer;

public class ZeroEvenOdd {

private int n, lastPrinted;

private Semaphore oddSem;

private Semaphore zeroSem;

private Semaphore evenSem;

public ZeroEvenOdd(int n) {

this.n = n;

this.lastPrinted = 0;

this.zeroSem = new Semaphore(1);

this.oddSem = new Semaphore(1);

this.evenSem = new Semaphore(1);

try {

this.evenSem.acquire();

this.oddSem.acquire();

} catch (InterruptedException e) {

throw new RuntimeException(e);

}

}

public void zero(IntConsumer printNumber) throws InterruptedException {

int numZeros = n;

while (numZeros-- > 0) {

this.zeroSem.acquire();

printNumber.accept(0);

if (lastPrinted % 2 == 0) {

this.oddSem.release();

} else {

this.evenSem.release();

}

}

}

public void even(IntConsumer printNumber) throws InterruptedException {

int numeven = n / 2;

while (numeven-- > 0) {

this.evenSem.acquire();

printNumber.accept(++lastPrinted);

this.zeroSem.release();

}

}

public void odd(IntConsumer printNumber) throws InterruptedException {

int numOdd = n - n / 2;

while (numOdd-- > 0) {

this.oddSem.acquire();

printNumber.accept(++lastPrinted);

this.zeroSem.release();

}

}

public static void main(String[] args) {

ZeroEvenOdd print = new ZeroEvenOdd(5);

Thread thread = new Thread() {

public void run() {

System.out.println("Thread Running");

IntConsumer ic = (x) -> System.out.println(x);

try {

print.zero(ic);

} catch (InterruptedException e) {

// TODO Auto-generated catch block

e.printStackTrace();

}

}

};

Thread thread2 = new Thread() {

public void run() {

System.out.println("Thread Running");

IntConsumer ic = (x) -> System.out.println(x +"\t");

try {

print.odd(ic);

} catch (InterruptedException e) {

// TODO Auto-generated catch block

e.printStackTrace();

}

}

};

Thread thread3 = new Thread() {

public void run() {

System.out.println("Thread Running");

IntConsumer ic = (x) -> System.out.println(x);

try {

print.even(ic);

} catch (InterruptedException e) {

// TODO Auto-generated catch block

e.printStackTrace();

}

}

};

thread.start();

thread2.start();

thread3.start();

}

}

Three threads started at the same time .Priniting Order should be maintained , it should always print first,second,third

 package printorder;

import java.util.concurrent.Semaphore;

import java.util.function.IntConsumer;

public class PrintOrder {

private Semaphore first;

private Semaphore second;

private Semaphore third;

public PrintOrder() {

this.first = new Semaphore(1);

this.second = new Semaphore(1);

this.third = new Semaphore(1);

try {

this.second.acquire();

this.third.acquire();

} catch (InterruptedException e) {

e.printStackTrace();

}

}

public void first() throws InterruptedException {

this.first.acquire();

print("first");

this.second.release();

}

private void print(String string) {

System.out.println("Print order :" + string);

}

public void second() throws InterruptedException {

this.second.acquire();

print("second");

this.third.release();

}

public void third() throws InterruptedException {

this.third.acquire();

print("third");

this.first.release();

}

public static void main(String[] args) {

PrintOrder order = new PrintOrder();

Thread t1 = new Thread() {

public void run() {

System.out.println("Thread Running t1 "+Thread.currentThread().getName());

try {

order.third();

} catch (InterruptedException e) {

// TODO Auto-generated catch block

e.printStackTrace();

}

}

};

Thread t2 = new Thread() {

public void run() {

System.out.println("Thread Running t2 "+Thread.currentThread().getName());

try {

order.second();

} catch (InterruptedException e) {

// TODO Auto-generated catch block

e.printStackTrace();

}

}

};

Thread t3 = new Thread() {

public void run() {

System.out.println("Thread Running t3 "+Thread.currentThread().getName());

try {

order.first();

} catch (InterruptedException e) {

// TODO Auto-generated catch block

e.printStackTrace();

}

}

};

t1.start();

t2.start();

t3.start();

}

}

Java Problems

Some Points on Priority queue

---------------------------------------

1)Since a priority queue needs to compare its elements and order them accordingly, the user defined class must implement the Comparable interface, or you must provide a Comparator while creating the priority queue. Otherwise, the priority queue will throw a ClassCastException when you add new objects to it.(https://www.callicoder.com/java-priority-queue/)

2)By default all wrapper class implement comparable interface

Leet Code problem 347. Top K Frequent Elements

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2

Output: [1,2]

Example 2:

Input: nums = [1], k = 1

Output: [1]

Solution we are going to use HashMap , Priority queue and Array

package LeetCode;

import java.util.HashMap;

import java.util.PriorityQueue;

import java.util.Queue;

public class TopKElements {

public int[] topKEmements(int[] nums, int k) {

HashMap<Integer, Integer> cache = new HashMap<>();

for (int num : nums) {

cache.put(num, cache.getOrDefault(num, 0) + 1);

}

Queue<Integer> minHeap = new PriorityQueue<>((a, b) -> cache.get(a) - cache.get(b));// Comparator so that it

// gets stored in the order

// where least occured

// element is stored in the

// top

for (int num : cache.keySet()) {

minHeap.add(num);

if (minHeap.size() > k) {

minHeap.poll();

}

}

int[] arr = new int[k];

int index = 0;

while (minHeap.size() != 0) {

int val = minHeap.poll();

arr[index++] = val;

}

return arr;

}

public static void main(String[] args) {

TopKElements elements = new TopKElements();

int[] nums = { 1, 1, 1, 2, 2, 3 };

int[] result = elements.topKEmements(nums, 2);

for (int number : result) {

System.out.println("Result is " + number);

}

}

}

Kth larets element in an array

Formula could be (length of array - k) this will give the index of the kth largest element

For example [ 2 1 3 8 9] is the array k =2 that means finf 2nd largest

Step 1: Sort the array [1 2 3 8 9]

step 2:length of array - k  --> (5 -2) which is 3 i.e at 3rd index we have 2nd largets element

Similarlly 3rd largets  means

5-3 which is 2 i.e at 2nd index we have 3rd largest element in the array

Using priority queue approach

----------------------------------------

1)Insert elements in to priority queue which will be inserted in to default ordering i.e least element will be removed first when we poll

2)We poll from queue when size becomes grater than k

that where we get result

Following is the code

package LeetCode;

import java.util.Arrays;

import java.util.PriorityQueue;

public class KthLargetsElement {

public int findkthLatgestUsingBruteForce(int[] numbers, int k) {

Arrays.sort(numbers);

int kthIndex = numbers.length - k;

return numbers[kthIndex];

}

public int findkthLargestUsingPriorityQueue(int[] numbers, int k) {

PriorityQueue<Integer> pq = new PriorityQueue<>();

for (int num : numbers) {

pq.offer(num);

if (pq.size() > k) {

pq.poll();

}

}

return pq.peek();

}

public static void main(String[] args) {

KthLargetsElement element = new KthLargetsElement();

int[] numbers = { 102 ,6,100,500,78 };

int k = 2;

int result = element.findkthLatgestUsingBruteForce(numbers, k);

System.out.println(k + "th largest element is " + result);

int resultUsingPriorityQueue = element.findkthLargestUsingPriorityQueue(numbers, k);

System.out.println("Result using priority queue " + resultUsingPriorityQueue);

}

}

String program Group Annagrams

----------------------------------------------

Input: strs = ["eat","tea","tan","ate","nat","bat"]

Output: [["bat"],["nat","tan"],["ate","eat","tea"]]

import java.util.ArrayList;

import java.util.Arrays;

import java.util.HashMap;

import java.util.List;

import java.util.Map;

public class GroupAnnagrams {

public List<String> groupAnagrams(String[] strs) {

// base case

if (strs.length == 0 || strs == null)

return new ArrayList<String>();

Map<String, List<String>> map = new HashMap<>();

for (String word : strs) {

// convert every string to a char array

char[] ch = word.toCharArray();

// sorted the array, this is easy for this logic to determine whether a key

// repeats or not

Arrays.sort(ch);

String sortedStr = String.valueOf(ch);

// if the hash map does not contain the sorted key in a given loop then we add

// it to the

// hash map along with a new ArrayList<String>() value

if (!map.containsKey(sortedStr))

map.put(sortedStr, new ArrayList<String>());

// we keep adding values to a list for a key in existing loop

map.get(sortedStr).add(word);

}

// return the group by iterating over all the values in the hash map and pass

// them to the constructor

// return new ArrayList<List<String>>(map.values());

return new ArrayList(map.values());

}

public static void main(String[] args) {

GroupAnnagrams grp = new GroupAnnagrams();

String[] strs = {"eat","tea","tan","ate","nat","bat"};

//String[] strs = { "" };

System.out.println("Annagrams are " + grp.groupAnagrams(strs));

}

}