Tip 1 : Make your fundamentals clear
Tip 2 : When you're Applying for the company try to do previous year question of that particular company
Tip 3 : Do at least 1 coding question per day
Tip 1 : Keep your resume clear and concise and it should be of 1 page only.
Tip 2 : In your resume lay down your project link and write down key points of the project instead of summary of the project.
It was medium level of question .Basically it had 3 coding questions . 1 from Dynamic Programming and other question from tree. I don't remember the other 2 questions.




As the answer can be very large, return the answer by taking modulo with 1000000007.
If ‘N’=1
So in 1 step we can reach either to ‘X’ , ‘Y’ or ‘Z’ and can not travel back to ‘O’.
Thus there are 0 ways.
If ‘N’ =2
So there are total three ways :
(i) O->X->O
(ii) O->Y->O
(iii) O->Z->O
If ‘N’ = 3
So there are total 6 ways :
(i) O->X->Y->O
(ii) O->X->Z->O
(iii) O->Y->X->O
(iv) O->Y->Z->O
(v) O->Z->X->O
(vi) O->Z->Y->O
A table T[][] is created where the row represents the number of ways and the column represents the position.
In order to fill the table, one observation needs to be made. That is, we can go back to the position O if we are not at O in the previous step.
Therefore, the number of ways to reach the origin O in the current step is equal to the sum of the number of ways the person is not at the origin O in the previous steps.
The round contain 2 coding question.



You may make as many transactions as you want but can not have more than one transaction at a time i.e, if you have the stock, you need to sell it first, and then only you can buy it again.
In this approach, we just need to find the next greater element and subtract it from the current element so that the difference keeps increasing until we reach a minimum. If the sequence is a decreasing sequence, so the maximum profit possible is 0.


Let the linked list be 1 -> 5 -> 2 -> 4 -> 3 -> 5 and then the last node is connected to node with value 4, which makes a loop. In this linked list, the loop starts from the node with value = 3 and ends at the node with value = 4.
You are supposed to break the loop ending at 4 and connect it to a node value just greater than 4, which is 5 in this case. So, the final answer is 1 -> 5 -> 2 -> 4 -> 3 and then the last node is connected to node with value 5.
The round contain coding question and simple data base query question. I don't remember the DB Question.



1. If ‘k’ is not present in 'arr', then print -1.
2. There are no duplicate elements present in 'arr'.
3. 'arr' can be rotated only in the right direction.
Input: 'arr' = [12, 15, 18, 2, 4] , 'k' = 2
Output: 3
Explanation:
If 'arr' = [12, 15, 18, 2, 4] and 'k' = 2, then the position at which 'k' is present in the array is 3 (0-indexed).
Compare X with the middle element.
If X matches with the middle element (arr[mid]), return the index mid.
If X is found to be greater than the arr[mid], then X can only lie in the subarray [mid + 1, end]. So search for X in the subarray {arr[mid + 1], .., arr[end]} .
Otherwise, search in the subarray {arr[start], …., arr[mid]}

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?