Tip 1 : Solve DSA problems if you don't have enough time for the interview on leetcode otherwise if you are in second year go for competitive programming.
Tip 2 : Be consistent try solve at least 1 problem each day and do up solving in contest.
Tip 3 : Don't forget other fundamental topics as they are important to like OOPS and MySQL.
Tip 4: I didn't have any significant project but try to do one which involves backend and frontend as it will be good in future aspects(after you join the company).
Tip 1 : If you don't have a good CGPA don't mention it in resume or do it at the end.
Tip 2 : Put on those things in resume which you are good in or you are sure about and highlight necessary points in resume as recruiter doesn't have time to read resume.
It was an online coding round which consisted of 20 mcq and 2 coding questions. 20 question were based on C++ i/o, DBMS, logic and aptitude and MySQL.



F(n) = F(n-1) + F(n-2),
Where, F(1) = F(2) = 1.
For ‘N’ = 5, the output will be 5.
This is a standard question where you have to print nth fibonacci number


An array 'C' is a subarray of array 'D' if it can be obtained by deletion of several elements(possibly zero) from the beginning and the end of array 'D'. For example, all the non-empty subarrays of array [1,2,3] are [1], [2], [3], [1,2], [2,3], [1,2,3].
Input: 'N' = 3 , 'ARR' = [-5, 10 , 0]
Output: -5
Explanation : The non empty subarrays possible for 'ARR' are [-5], [10], [0], [-5, 10], [-5, 0], [10, 0], [-5, 10, 0]. The sum of the elements of these subarrays are -5, 10, 0, 5, -5, 10, 5. The minimum of them is -5.
Idea is to use kadane's algorithms where you are taking prefix sum of the array but if prefix sum becomes negative then set it to zero while doing it max prefix sum that you get in the array is the answer.
Interview was at morning 12 pm and interviewer was pretty cool and helpful throughout the process. Whenever I got stuck he game me some hint about that problem.



In cases where the dividend is not perfectly divisible by the divisor, you are required to return the integer value of the quotient which is closer to zero.
For example - If the answer is '8.34', we return the integer part, i.e., '8'. Also, If the answer is '-2.67', we return the integer part, i.e., '-2'.
Assume we're dealing with a system that can only store integers in the 32-bit signed integer range: [2^31, 2^31-1]. If the quotient is higher than 2^31 - 1, return 2^31 - 1; if it is less than -2^31, return -2^31.
If the answer is ‘9.333’, then return ‘9’, or if the answer is ‘-8.123’, then return ‘-8’.
This can be solved using bit manipulation using left shift .When we apply left shift operator to an integer it gets multiplied by 2. So I keep divisor multiplying by 2 until it become greater than the dividend and then again one time divide it by 2. After that i subtract divisor from dividend. and make the subtracted number divisor and apply the same process again until the dividend becomes less than divisor.


HR round
Introduce yourself.
Why do you want to join us?

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