Interview Questions

Inraathp R. - "Clarify : "Movie watching” can happen in a Theatre On a cable TV via an OTT app, A dedicated home-theatre set-up. I will focus on OTT streaming on a Smart-TV because: It is the fastest-growing surface (70 % of long-form streaming minutes). Geography: USA Platform : Any (Android TV, Tizen etc) OTT: Netflix What are we trying to improve ? Basic Journey -->Switch on TV->Navigate to OTT->Find Movie->Play Movie->Consume Audio and Video Setup Journey -> Create account -> Setu"See full answer

Natalie C. - "I walked through the code for a react.js based tic-tac-toe game written in typescript. The goal was to find ways to improve the code/ suggest improvements. I missed some areas like where state was being updated directly rather than using React's setState. There were issues around clear and maintainable logic, adherence to React best practices."See full answer

Jiin S. - "I would use A/B testing to see if the new feature would be incrementally beneficial. To begin the testing, we should define what's the goal of this testing. Let's say the new feature would increase the average number of trade by X. Then randomly assign the clients to two groups, control and test group. Control group doesn't see the new feature and the test group see the new feature. We could also stratified sampling if we want to make sure cover different customer segmentation. During this desig"See full answer

Abhishek M. - "System Overview and Workflow: The architecture is composed of distributed microservices that communicate via an event bus (Apache Kafka). Ingestion: The process begins when the Chargeback Ingestion Service receives dispute notifications from acquiring banks through various methods (webhooks, SFTP files) and standardizes them into a canonical format. Lifecycle Management: A DisputeNotification event is published to Kafka, which the Chargeback Lifecycle Service consumes. This service acts"See full answer

Mohammed H. - "First i will collect relevant data,Actual problem,Necessary input and raw material then i will start design drawing,modeling,Callibration and finally handing over my work."See full answer

Payal B. - "My brute force approach was to read them. Give a id to each paragraph and for each token count the number of time it has appeared. If any two rows look same , it is duplicated. Further , interviewer guided me that he will do it with hashing."See full answer

Nils G. - "Bitshift the number to the right and keep track of the 1's you encounter. If you bitshift it completely and only encounter one 1, it is a power of two."See full answer

VContaineers - " Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between: Skipping the current house and taking the maximum money stolen up to the previous house. Robbing the current house and adding its value to the maximum money stolen up to the house two steps back. package main import ( "fmt" ) // rob function calculates the maximum money a robber can steal func maxRob(nums []int) int { ln"See full answer

Asish B. - "Number of employees after the first year = n*(1+r) = n1 Number of employees after the second year = n1(1+r) = n(1+r)**2 Hence, the number of employees after 't' years = n(1+r)*t"See full answer

Frederick A. - "This problem can be solved with two approaches Iterative approach Recursive approach Quite easy to think about the iterative approach, you can make use of a while loop in that case. But what if you want to make use of previously computed values? That case going for the recursive solution is quite useful. class Collatz: def init(self) -> None: self.cache = {} self.steps = 0 def steps_from(self, n) -> int: # base case if n == 1: "See full answer

Anonymous Wolf - "Construct a min-heap either inplace, or by making a copy of the array and then applying heapify on that copy. This is done in O(n) time. Maintain two zero-initialised variables - sum and count. Keep popping off the heap while sum < k, and update count. In the worst case you will have to do n pops, and each pop is O(log n), so the algorithm would take O(n log n) in total. Space complexity depends on whether you're allowed to modify inplace or not, so either O(1) or O(n) respectively."See full answer