Interview Questions

John F. - "First, I would like to establish some assumptions about the problem to determine if this is possible. For this scenario, I will assume that “accepted” does not mean a student is enrolled at the university (most students apply to multiple universities and can only attend one university). For simplicity, I will also assume that each student is applies to a single department and that every student is associated with a single department. Based on this information, I see a scenario that this would"See full answer

VictorSage - "The critical data to forecast projects with greater accuracy are Remaining Estimated work, Customer needed date, Cost of Delay."See full answer

Ekta M. - " Why do we want to increase the price? How much do we need to increase? Is it tied to some kind of revenue target? What is the region in which price needs to be increased? - Say India Who are the major players in that region who offer similar services and how are they priced? Will the users switch to those competitors? What is our unique proposition for retention? Do we have any leverage against them which will stop the users from churning? Any change in pricing or anything major like terms an"See full answer

Yizhi T. - "bidirectional tranverse - the nearest seller is either on the left or right of each child 1 - tranverse from left to right，always record the latest seller si, and record the left nearest distance ki - si of current kid into leftarray 2 tranverse from right to left，always record the latest seller si, and record the left nearest distance si-ki of current kid into rightarray 3 find the maximum of the min(leftarray[i], rightarray[i]) of each kid Time complexity: O(N+M) 3 passes of tra"See full answer

Anonymous Unicorn - "naive solution: def countprefixpairs(words): n = len(words) count = 0 for i in range(n): for j in range(i + 1, n): if words[i].startswith(words[j]) or words[j].startswith(words[i]): count += 1 return count using tries for when the list of words is very long: from collections import Counter class TrieNode: def init(self): self.children = {} self.count = 0 # To count the number of words ending at this node"See full answer

Date G. - "Lets start with assumptions 100m people go out to bars in US Assuming COVID is behind us Also assuming this is annually and not monthly or daily (though it might be better to ask this) Step 1: Why should Google build something like this Google Mission: Make information accessible and useful Google's current state: Focus on improving current products (improve engagement and thus monetization) while also building things for future (growth) Product Mission: Make going to bar a"See full answer

Hl M. - "Clarifying Slack is a mature product in markets with millions of users B2B platform Expand customer base to school boards SWOT Technology is mature and well reputed Good privacy practices Threats Google school Microsoft learning education Zoom Journeys User Users for slack - for - school School boards 12 school boards Primary Secondary High Universities - Remote learning / Distant educations - Incumbents MooCs - Existing platform End Users: Students Teachers Administrators Ed"See full answer

onering2ruleall - "Total volume of the swimming pool, let's say it's a standard 50m x 25 m x 2m =2500 m3 = 2.5 million litter standard water horse water flow rate is 17 liter per min. So, 2.5 m / 17 =147,059 min"See full answer

Anubhav A. - "Clarifying question: Do we want to judge need of an existing product or an idea that can be converted into a product Let me solve for both the cases. Q: Determine if there is a need for a platform that is already live and has some users. While thinking about need, I am finding it resonating with "Does this product have a pmf". Few ways to do that: Take a look at the retention number (D1, D3, D7 and D15). Basis on the product the time range can be different but if the retention number is"See full answer

Babaa - "Function signature for reference: def calculate(servers: List[int], k: int) -> int: ... To resolve this, you can use binary search considering left=0 and right=max(servers) * k so Example: servers=[1,4,5] First server handle 1 request in let's say 1 second, second 4 seconds and last 5 seconds. k=10 So I want to know the minimal time to process 10 requests Get the mid for timeline mid = (left+right)//2 -> mid is 25 Check how many we could process 25//1 = 25 25//4=6 25//5=5 so 25 + 6 +"See full answer

Amparo L. - "Yes I do, we discuss our agendas as a team and then we work together to get the tasks done before our work shift ends."See full answer