Core%20Java%20Interview%20Questions%20and%20Answers

Question: How HashMap works in Java?

Answer: "How does get () method of HashMap works in Java"

And then you get answers like I don't bother its standard Java API, you better look code on java; I can find it out in Google at any time etc.
But some interviewee definitely answer this and will say "HashMap works on principle of hashing, we have put () and get () method for storing and retrieving data from hashMap. When we pass an object to put () method to store it on hashMap, hashMap implementation calls
hashcode() method hashMap key object and by applying that hashcode on its own hashing funtion it identifies a bucket location for storing value object , important part here is HashMap stores both key+value in bucket which is essential to understand the retrieving logic. if people fails to recognize this and say it only stores Value in the bucket they will fail to explain the retrieving logic of any object stored in HashMap . This answer is very much acceptable and does make sense that interviewee has fair bit of knowledge how hashing works and how HashMap works in Java.
But this is just start of story and going forward when depth increases a little bit and when you put interviewee on scenarios every java developers faced day by day basis. So next question would be more likely about collision detection and collision resolution in Java HashMap ->

"What will happen if two different objects have same hashcode?"

Now from here confusion starts some time interviewer will say that since Hashcode is equal objects are equal and HashMap will throw exception or not store it again etc. then you might want to remind them about equals and hashCode() contract that two unequal object in Java very much can have equal hashcode. Some will give up at this point and some will move ahead and say "Since hashcode () is same, bucket location would be same and collision occurs in hashMap, Since HashMap use a linked list to store in bucket, value object will be stored in next node of linked list." great this answer make sense to me though there could be some other collision resolution methods available this is simplest and HashMap does follow this.

"How will you retreive if two different objects have same hashcode?"


Interviewee will say we will call get() method and then HashMap uses keys hashcode to find out bucket location and retrieves object but then you need to remind him that there are two objects are stored in same bucket , so they will say about traversal in linked list until we find the value object , then you ask how do you identify value object because you don't value object to compare ,So until they know that HashMap stores both Key and Value in linked list node they won't be able to resolve this issue and will try and fail.

But those bunch of people who remember this key information will say that after finding bucket location , we will call keys.equals() method to identify correct node in linked list and return associated value object for that key in Java HashMap. Perfect this is the correct answer.

In many cases interviewee fails at this stage because they get confused between hashcode () and equals () and keys and values object in hashMap which is pretty obvious because they are dealing with the hashcode () in all previous questions and equals () come in picture only in case of retrieving value object from HashMap.
Some good developer point out here that using immutable, final object with proper equals () and hashcode () implementation would act as perfect Java HashMap keys and improve performance of Java hashMap by reducing collision. Immutability also allows caching there hashcode of different keys which makes overall retrieval process very fast and suggest that String and various wrapper classes e.g Integer provided by Java Collection API are very good HashMap keys.

Now if you clear all this java hashmap interview question you will be surprised by this very interesting question "What happens On HashMap in Java if the size of the Hashmap exceeds a given threshold defined by load factor ?". Until you know how hashmap works exactly you won't be able to answer this question.
if the size of the map exceeds a given threshold defined by load-factor e.g. if load factor is .75 it will act to re-size the map once it filled 75%. Java Hashmap does that by creating another new bucket array of size twice of previous size of hashmap, and then start putting every old element into that new bucket array and this process is called rehashing because it also applies hash function to find new bucket location.

If you manage to answer this question on hashmap in java you will be greeted by "do you see any problem with resizing of hashmap in Java" , you might not be able to pick the context and then he will try to give you hint about multiple thread accessing the java hashmap and potentially looking for race condition on HashMap in Java.

So the answer is Yes there is potential race condition exists while resizing hashmap in Java, if two thread at the same time found that now Java Hashmap needs resizing and they both try to resizing. on the process of resizing of hashmap in Java , the element in bucket which is stored in linked list get reversed in order during there migration to new bucket because java hashmap doesn't append the new element at tail instead it append new element at head to avoid tail traversing. if race condition happens then you will end up with an infinite loop. though this point you can potentially argue that what the hell makes you think to use HashMap in multi-threaded environment to interviewer.