Developer Interview Questions in Gurgaon, Haryana

What is overriding in java?

Polymorphism: - Polymorphism is a greek word, when we break this word

Polymorphism is a greek word when we break this word into two parts like poly and morph it means poly says "Many" and morph means "meanings" , so one form having different meanings are said to be polymorphism. Eg. Break system in vehicles. Overriding in Java:-- See if the two methods are there one is parent and other is child then child able to access the methods arguments through inheritance but some of the arguments not satisfied by the child class then it will make some changes by its own way and will adding new argumets which is suitable for child class that process is nothing but overriding.

use TreeMap for characters frequency and print top k keys.

do below steps:- 1) get the middle node of linked list and total number of nodes in O(n) . 2) get a stack of size n/2 3) start pushing node values of linked list to stack till(n-1/2 th node if n is odd else n/2 th node) 4) pop an element from stack and compare with next remaining node of the linked list after middle node. 5) if all values are same till stack is empty then palindrome else not.

Solution with O(n) time, and O(1) space complexity: 1) Use two pointers to find the middle node (i.e, define pointer A, B and initialise them to head. Move pointer by one step, and B by two steps. When B reaches the null, A will be in the middle of the list). The n/2 approach takes extra iteration to compute the size/length of the list. 2) Now reverse the first half of the list. 3) Compare first and second half one by one. If there is a mismatch, the list is not palindrome.

*Armstrong Number= { int num=370,number,temp,total=0; number=num; while(number!=0) { temp=number%10; total=total+temp*temp*temp; number/=10; } if(tota==num) System.out.println(num+"is an Armstrong Number"); else System.out.println(num+"is not an Armstrong Number"); } *Factorial using For Loop= { int number=5; long fact=1; for(int i=1;i<=number;i++) { fact=fact*i; } System.out.println("Factorial of "+number+" is:"+fact); } *Factorial using While Loop= { int=5; long fact=1 int i=1; while(int i<=number) { fact=fact*i; i++; } System.out.println("Factorial of "+number+" is:"+fact); }

*Armstrong Number= { int num=370,number,temp,total=0; number=num; while(number!=0) { temp=number%10; total=total+temp*temp*temp; number/=10; } if(tota==num) System.out.println(num+"is an Armstrong Number"); else System.out.println(num+"is not an Armstrong Number"); } *Factorial using For Loop= { int number=5; long fact=1; for(int i=1;i<=number;i++) { fact=fact*i; } System.out.println("Factorial of "+number+" is:"+fact); } *Factorial using While Loop= { int=5; long fact=1 int i=1; while(int i<=number) { fact=fact*i; i++; } System.out.println("Factorial of "+number+" is:"+fact); }

*Armstrong Number= { int num=370,number,temp,total=0; number=num; while(number!=0) { temp=number%10; total=total+temp*temp*temp; number/=10; } if(tota==num) System.out.println(num+"is an Armstrong Number"); else System.out.println(num+"is not an Armstrong Number"); } *Factorial using For Loop= { int number=5; long fact=1; for(int i=1;i<=number;i++) { fact=fact*i; } System.out.println("Factorial of "+number+" is:"+fact); } *Factorial using While Loop= { int=5; long fact=1 int i=1; while(int i<=number) { fact=fact*i; i++; } System.out.println("Factorial of "+number+" is:"+fact); }

The coding problem statement was too complex and the peer programmer was not even ready to explain the scenarios to be handled. I had written the code which was working with inputs provided, approach was also correct, wrote unit tests as well. But the peer programmer rejected it without giving any feedback. I believe either thought works is recruiting for candidates to build some robot/missiles or they does not want to recruit anyone and just show the world that they are recruiting in COVID situation. Was not expecting this from Thought works. Very Disappointing Experience.

Irony is even peer programmer don't know how to do code pairing. Same happened with me last year and I was very disappointed from peer response.

What was the problem statement?

I have done my all questions easily so i haven't face any diificulty in Technical interview.

Do you remember the topics for aptitude round ?

Actually there is no particular syllabus for aptitude but do the questions for which you are confident that ans is correct and do attempt only if you are sure that you can make an success full attempt. I have done the questions more based on logical reasoning because for me they are doable and less time taking and i am sure about them. Do not think to do more attempts but you have to accurate. I guess i have attempted only 12 questions out of 20 and got shortlisted for next round.

Ans. I answered: Temporary tables are stored only in one database i.e. tempdb. Still the interviewer was asking irrelevant scenarios.

No, because temporary table only exists in temp db

Interviewer asked, "but if temporary table exists in different databases, can both be accessed in same query window?"

We can use a HashMap for it where name will be the key and phone number will be the value. If second time any name appears then its key will be already present in the map and it will be deleted.

Name can be same. A contact will be duplicated only if phone number is same. Need to use number as key.

Instead of map you can use set