Tip 1 : Practice all the standard interview problems of every topic which will count upto approx 400 questions.
Tip 2 : Do atleast two good projects and also prepare them so that you can answer all the cross questions asked related to it.
Tip 3 : Study CS Fundamentals and OOPS, don't take it lightly.
Tip 1 : Make one page resume and add atleast two projects and your coding profiles skills and achievements.
Tip 2 : Do no fake out your resume it will fire back at the time of interview.
There were 10 MCQ questions based on CS fundamentals and OOPS and 2 coding questions based on data structures and algorithms. It was on HackerEarth platform and the timing of the test was from 4 pm to 6.30 pm there was no slot. The camera and tab change proctoring were there.


Input: ‘arr’ = [1, 1, 2] and ‘k’ = 2
Output: 2
Explanation: If we want to make two subarrays, there are two possibilities: [[1], [1, 2]] and [[1, 1], [2]]. We can see that the maximum sum of any subarray is minimized in the second case. Hence, the answer is 2, which is the maximum sum of any subarray in [[1, 1], [2]].


If X = 6
If all three moves are taken to the right then the robot will end up at 1 + 2 + 3 = 6, so a minimum of three moves are needed.
The round was scheduled around 10 AM in the morning and the interviewer was quite friendly and cooperative in nature and also gave hints of the coding problem asked by him, the interview was taken on google meet and coding question was explained on google docs. The interview lasted for 60 minutes.


You are given preOrder = [10, 5, 1, 7, 40, 50], the binary search tree from the given preorder traversal is

Hence the answer is [1, 7, 5, 50, 40, 10].
Intution
Since it is preorder traversal of bst the first element will be root node and all the smaller than root node will be in the left subtree and all the larger once will be in the right subtree.So we will use the above logic and construct the tree recursively and the time complexity will be O(N).
We follow the following steps
1> we create the node
2> we traverse the array for values that are less than the current node these values will become our left subtree. we stop whenever we get a value larger than the current root of the subtree.
3> we take the rest of the array (values that are greater than the value of the current root)-these are the values that will make out the right subtree!
so recursively we make a root!
make the left subtree(recursively)
then make right subtree(recursively)
Difference between process and thread.
Scheduling Algorithms.
Deadlocks.
Semaphore and Mutex.
Static and Dynamic Binding
Page fault and Demand Paging.
Tip 1 : Prepare well for OS from an article or youtube video.
Tip 2 : Go through the most asked OS questions articles available on the internet.
Tip 3 : since this is a theoretical subject revise thoroughly.
Normalization and its types.
Transactions.
ACID properties of DBMS.
Tip 1 : Prepare well for DBMS from an article or youtube video.
Tip 2 : Go through the most asked DBMS questions articles available on the internet.
Tip 3 : since this is a theoretical subject revise thoroughly.
The round was scheduled around 12 PM in the afternoon and the interviewer was quite friendly and cooperative in nature and also gave hints of the coding problem asked by him, the interview was taken on google meet and coding question was explained on google docs. The interview lasted for 60 minutes.


step 1 – I first gave him a brute force approach like checking for all subarray and taking the maximum sum, it was a O(N^2) approach.
step 2 -- then I moved to optimized approach and told him the pretty standard kadane algorithm which is a O(N) time complexity and also I was able to code it quickly.




1. You are not required to print the output explicitly, it has already been taken care of. Just implement the function and return the ‘K-th’ smallest element of BST.
2. You don’t need to return ‘K-th’ smallest node, return just value of that node.
3. If ‘K-th’ smallest element is not present in BST then return -1.
Step 1 -- since it is a bst and preorder traversel of bst is in sorted order so by using an extra space we can easily store the preorder traversal of the bst and return the kth element .
Now he told me to do it without using extra space. So for this optimization
Step 2 -- the simplest logic is to do reverse inorder traversal and while doing reverse inorder traversal simply keep a count of the number of Nodes visited. When the count becomes equal to k, we stop the traversal and print the data. It uses the fact that reverse inorder traversal will give us a list sorted in descending order.
the timing of the round was approx 3 PM and it was conducted on google meet it was just more like a discussion round and some standard hr questions.the interviewer seemed very polite and humble.
The interviewer greeted me and asked me how was my day.
It was more like a discussion rather than an Interview.
she started with the standard question, Tell me about yourself.
I had already prepared answers for 2-3 types of HR round introduction and Behavioral questions. So I was able to give him a short, precise, and to-the-point Introduction of mine.
She also asked me how much I am aware of the company like what the company does etc.
Then she moved on to the Projects that I had mentioned on my resume,
I told about 1-2 Projects which I did until now and briefly explained all of them.
Then she asked me if I had any questions for him.
I asked a few questions, and she was happy to answer.
Tip 1 : Prepare some standard HR questions available on various websites.
Tip 2 : revise your projects mentioned in the resume.
Tip 3 : Do proper communication and be confident throughout the discussion and be honest about every answer.

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