Tip 1 : Having a prior internship with a dev role will definitely help in interviews.
Tip 2 : If you are not good in frontend then prepare DSA and computer fundamentals very well
Tip 3 : Apart from Leetcode medium questions Have hands on DMBS queries
Tip 4 : Having an API based project will be very helpful for backend roles
Tip 1 : They will mostly ask in first round if you interested in frontend or backend roles and then the follow up interviews will happen in that direction.
Tip 2 : so make your resume specific to the role you are interested in.
Tip 3 : Write what you have worked on in your prior internship.
Tip 4 : if this is your first internship write your strong points and only things which you are confident in because after selection interviewer will driil you on those topics.
three coding questions , no mcq, 1 hr limit



The given linked list is 1 -> 2 -> 3 -> 4-> NULL. Then the reverse linked list is 4 -> 3 -> 2 -> 1 -> NULL and the head of the reversed linked list will be 4.
Can you solve this problem in O(N) time and O(1) space complexity?



1. You can return the list of values in any order. For example, if a valid triplet is {1, 2, -3}, then {2, -3, 1}, {-3, 2, 1} etc is also valid triplet. Also, the ordering of different triplets can be random i.e if there are more than one valid triplets, you can return them in any order.
2. The elements in the array need not be distinct.
3. If no such triplet is present in the array, then return an empty list, and the output printed for such a test case will be "-1".
sort the list in ascending order then find the complement after that do a two pointer approach with a front and rear pointer to find the target value and add the numbers result vector and ensure there are no duplicates by: check if the the number is already in our list, if so move the pointers



The Linked Lists, where a1, a2, c1, c2, c3 is the first linked list and b1, b2, b3, c1, c2, c3 is the second linked list, merging at node c1.

1st they asked area of interest like frontend or backend. I said backend
Then thy asked a array based question followed by a linked list.
Then he asked some computer fundametals.



1. All the elements are in the range 0 to N - 1.
2. The elements may not be in sorted order.
3. You can return the duplicate elements in any order.
4. If there are no duplicates present then return an empty array.
If the number is visited it is turned negative
If the visited number is negative that means it has been already visited(duplicate) and hence added to the list.
Since the array values are from 1 to n therefore the 1 is subtracted from the array values at the time of indexing.




basic idea is to have two pointers, one for previous node and one for current, if current node.val == val
we want to set our previous.next to current.next.
mainly focused on dsa , dbms and working of an API



If the given list is {1 -> -2 -> 3} (which is sorted on absolute value), the returned list should be {-2 -> 1 -> 3}.
Use 2 pointers: fast and slow to divide the list into 2 sublist: list1 and list2 and make sure list1 is equal to or is longer than list2.
The key is the condition of while loop while(fast.next!=null && fast.next.next!=null). After this while loop slow will be at the position of the end of list1.
sort list1 and list2
merge list1 and list2
this was a hr round who asked about personal experiences , expectations, and behavioural questions.

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
What is recursion?