Software Engineer Interview Questions

GalacticInterviewer - "I first clarified what he needed for the leaderboard meaning did he need a gaming leaderboard system, a stock market leaderboard system where the trades placed would be ranked in terms of values or a coding platform leaderboard where the users are ranked as per the points earned by solving the problem? He answered that he wanted a multipurpose leaderboard system He was specifically interested in the design of the api part Explained him that assuming we had a flask api there would be a get re"See full answer

Rahul R. - "I would first call out the two types of Google Doodles that are published. First would be the static doodle which just shows a picture, and users can get more information on it by hovering on it or clicking on it. Second would be the dynamic google doodles which are either short videos or games through which users engage a bit more. My top 3 metrics that covers both would be: Click-through rate and post-click CTR - Although different metrics, I clubbed them since they fall under similar"See full answer

Dinesh K. - "i dont know"See full answer

Vaishnavi I. - "create models -> truck, driver, trip,status. trip will have foreign key relationships with truck,driver and status. Create CRUD endpoints for all the models. Create a read by truck id endpoint that filters the data by id (can also filter the data from the previous mentioned endpoint). update trip data can also update the driver status of the truck."See full answer

Anonymous Starfish - "Ask my manager how urgent is my task and why it should be done if needed. If its urgent enough and I can’t wait, communicate with team explaining what goals going to be achieved by implementing and why its important"See full answer

Amparo L. - "One Accomplishment I'm most proud of is that I graduated from Schaumburg High School In May of 2021 and I was able to get up the stage and collect my diploma. This was a HUGE Impact in regards of passing all of my classes and earning all of my credits in order to be apart of the NOW Arena graduation."See full answer

Aditya S. - "I told a story about how our team was focussed on moving a key metric i.e. NPS and to do that we build 3 top requested user feature. Post release the detractors % didn't move even though the detractors request for features shipped went down. Then I connect with users and did some analysis post which we realised that we need to pivot our focus from shipping features to enabling complete workflows for our users i.e. shipping all those feature which are used together in a feature as then only users"See full answer

Tiago R. - "class Node { constructor(data) { this.data = data; this.left = null; this.right = null; } } function exploreSubtree(node) { let leftHeight = 0; let rightHeight = 0; let maxDiameter = 0; // visit left if (node.left) { const leftVisit = exploreSubtree(node.left); leftHeight = leftVisit.height + 1; maxDiameter = Math.max(maxDiameter, leftVisit.maxDiameter); } // visit right if (node.right) { con"See full answer

Tiago R. - "function serialize(list) { for (let i=0; i 0xFFFF) { throw new Exception(String ${list[i]} is too long!); } const prefix = String.fromCharCode(length); list[i] = ${prefix}${list[i]}; console.log(list[i]) } return list.join(''); } function deserialize(s) { let i=0; const length = s.length; const output = []; while (i < length) { "See full answer

Anonymous Unicorn - "python: def justifywords(wordslist, width): result = [] currlinechar_count = 0 curr_words = [] for word in words_list: if curr_words: space_needed = len(word) + 1 # Space needed for the word and a preceding space else: space_needed = len(word) if currlinecharcount + spaceneeded > width: result.append(' '.join(curr_words)) curr_words = [word] currlinechar_count = len("See full answer

Shuchi A. - "A few months ago I joined a micro-services platform engineering team as their manager, at that time my team was struggling to deliver towards an upcoming production deadline for a customer facing product. Production date had been moved 5 times already and there were about 40% of product features which were remaining to be tested and signed off to move to production . I was made responsible to deliver the release of this product within the deadline and turnaround the software delivery throughput."See full answer

Vikash A. - "Functional Requirement Upload the file of any type POST v1/drive Upload the another version of file. POST v1/drive/{fileId} Delete the file. DELETE v1/drive/{fileId} Share the file. POST v1/drive/{fileId} Control the access level of the file. Provide the file accessibility following the directory structure. Non Functional Requirement Reliability: The file along with its versions uploaded should be"See full answer

Scott S. - " Project Overview: Real-Time Risk Management System Objective The goal was to develop a real-time risk management system capable of processing and analyzing large volumes of trading data to provide near-instantaneous risk assessments. This system was crucial for enabling traders to make informed decisions while managing their exposure to various market risks in real-time. Complexity Factors 1. \\Data Volume and Velocity\\ \\High Throughput:\\ The system needed to ha"See full answer

Matthew K. - " function climbStairs(n) { // 4 iterations of Dynamic Programming solutions: // Step 1: Recursive: // if (n <= 2) return n // return climbStairs(n-1) + climbStairs(n-2) // Step 2: Top-down Memoization // const memo = {0:0, 1:1, 2:2} // function f(x) { // if (x in memo) return memo[x] // memo[x] = f(x-1) + f(x-2) // return memo[x] // } // return f(n) // Step 3: Bottom-up Tabulation // const tab = [0,1,2] // f"See full answer

Batman X. - "def friend_distance(friends, userA, userB): step = 0 total_neighs = set() llen = len(total_neighs) total_neighs.add(userB) while len(total_neighs)!=llen: s = set() step += 1 llen = len(total_neighs) for el in total_neighs: nes = neighbours(friends, userA, el) if userA in nes: return step for p in nes: s.add(p) for el in s: total_neighs.add(el) return -1 def neighbours(A,n1, n2): out = set() for i in range(len(A[n2])): if An2: out.add(i) return out"See full answer