Tip 1 : Practice machine coding questions from various platforms
Tip 2 : Practice standard interview questions from internet, cover all data structures and important algorithms
Tip 3 : Note down your approach somewhere and revisit that document later to revise
Tip 1 : Use standard templates, don't go for fancy ones
Tip 2 : Keeps the info short and concise with important keywords
It was a machine coding round.
Interviewer will explain and present a low level design question.
90 minutes to prepare design and code without using any databases.
After this we have a one on one with a panel to discuss the solution.
Implement an e-commerce platform application.
Tip 1 : Read the requirements doc thoroughly and come up with a small class diagram on paper before starting to code
Tip 2 : Make sure to test the code after every new feature if possible
Tip 3 : Make sure to cover corner cases
This is a Problem solving round where they ask two DSA questions. We have to discuss possible solutions and write pseudo code for them. Interviewers were friendly and polite.



1) Yuki can buy 1 more blade with cost 'A.’ He now has ‘K+1’ Ninja blades.
2) Yuki could buy a ‘K’ number of blades with cost 'B.’ He now has ‘2*K’ blades.
where 'A' and 'B' are predefined and constant.
There can be two or more ways with the exact cost. You can consider any one of them, but the overall cost to reach from 0 to 'N' must be minimized.
Consider Yuki need to buy 5 blades, the cost of adding 1 blade is 2, and the cost of doubling the blades is 1 then you have to perform the following operations:
1) Doubling 0 will result in 0 only, so add 1 blade to 0 blades with cost 2. Total cost becomes 2.
2) Next, you can either double 1 to reach 2 or add 1 blade. But since the cost of doubling is less than that of adding, so double 1 with cost 1. Total cost becomes 3.
3) Doubling 2 will result in 4 with a cost of 1. Total becomes 4.
4) Adding 1 in 4 will result in 5 (which is the desired number) with a cost of 2. The total cost to reach 5 becomes 6.
If we fix j, then i and k are independent. We can get the best i by iterating while i
We can do the same for j
Time complexity O(n^2)



1
/ \
2 3
The root to leaf path 1->2 represents the number 12.
The root to leaf path 1->3 represents the number 13.
The total sum of all the possible root to leaf paths is 12+13 = 25
The output may be very large, return the answer after taking modulus with (10^9+7).
We need to check recursively if the node to leaf path sum of a node's left or right child is equal to (target - current node value)
Interviewer asked me to give all such paths. I suggested using a stack to store the path, pop nodes off the stack recursively if the path sum does not add up. If it does add up then store the contents of the stack as a path in a list of lists. Finally return this list.
It was a hiring manager round. The HM asked me to introduce myself and asked about my background, work experience. It was a 45 -50 minutes round with a senior engineering manager.
Tip 1 : Keeps things honest and simple
Tip 2 : Make sure you don't contradict yourself
Tip 3 : Speak confidently

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?