Tip 1 : Look out for the behavioural questions
Tip 2 : Be ready to write production-level code (No edge case misses in your final code)
Tip 3 : Make sure to dictate and walk through the solution instead of blatantly writing code
Tip 1 : Mention previous experiences with individual contributions
Tip 2 : Include projects with high impacts if low in work experience
2 DSA questions along with some debugging questions



You had to make greedy choices at each step and find the optimal solution



Note: Since the number of ways can be very large, return the answer modulo 1000000007.
N=3

We can climb one step at a time i.e. {(0, 1) ,(1, 2),(2,3)} or we can climb the first two-step and then one step i.e. {(0,2),(1, 3)} or we can climb first one step and then two step i.e. {(0,1), (1,3)}.
It was either of the 3 choices for each step and could be easily solved by using DP and memoising optimal ans for a state to compute ones in the future.
1 Behavorial question + 2 DSA questions



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 array and use the 2 pointer technique to find triplets that sum up to a given (X) -> expected Time complexity -> O(N^2)



The given linked list is 1 -> 2 -> 3 -> 2-> 1-> NULL.
It is a palindrome linked list because the given linked list has the same order of elements when traversed forwards and backward.
Can you solve the problem in O(N) time complexity and O(1) space complexity iteratively?
Solved it in 2 steps
1) Find the middle of the Linked List (Standard problem)
2) Reverse the second hald of the Linked List(Standard problem)
Compare both havles and return the result
1 Behavorial question + 2 questions on DSA



Gave the approach that dist(a,b) = distFromRoot(a) + distFromRoot(b) - 2*distFromRoot(LCA(a,b))




Consider the above binary tree. If ‘K’ is 15 then the required paths are [5, 6, 4] and [5, 15, -5].
Could only give the approach as was low on time, that we could run a traversal from each node and generate all paths from a specific node to check if it had an average of X and appeand to the answer if the case.
O(N^2)
Past work discussion + 1 DSA question



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.
Just a simple hashmap solution or a sorting based solution
1) Discussion on previous work
2) Largest impact you have had in your previous organisation
Had a chat , was thrown some counter questions and addressed them

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
How do you remove whitespace from the start of a string?