Recent Software Engineer Interview Questions

Anonymous Wasp - "I was able to provide the optimal approach and coded it up"See full answer

Luke P. - "A full stack developer could be summarized as the person who both writes the APIs and consumes the APIs. They are familiar with Databases/Data-layer services, middle-layer/application services and business logic, and finally familiar with the consumers whether front-end applications/UIs or other systems. They can understand the trade-offs up and down the stack, where to adjust along the service-call-path. Ideally they are comfortable programming both async calls (front end javascript promises, e"See full answer

Ravi K. - "select e.firstname as firstname, m.salary as manager_salary from employees e join employees m on e.manager_id = m.id where e.salary > m.salary; `"See full answer

Simon O. - "Functional Requirement Ingest Book Reviews in real time User in the website needs specify title tho search books, return response is the list of reviews on the books. Non Functional Requirement: User get real-time book reviews> It nees to handle the search among 1 million of active users daily"See full answer

叶 路. - "from collections import deque def updateword(words, startword, end_word): if end_word not in words: return None # Early exit if end_word is not in the dictionary queue = deque([(start_word, 0)]) # (word, steps) visited = set([start_word]) # Keep track of visited words while queue: word, steps = queue.popleft() if word == end_word: return steps # Found the target word, return steps for i in range(len(word)): "See full answer

Aman G. - "Was given 90 minutes with an exhaustive set of requirements to be implemented as a full-stack coding exercise. It was supposed to have a UX, a server and a database to store and retrieve data. The IDE was supposed to be self-setup before the interview. The panel asked questions on top of the implementation around decision making from a technical perspective"See full answer

Tes d H. - "Here is my first shot at it. Please excuse formatting. To find the maximum depth of the dependencies given a list of nodes, each having a unique string id and a list of subnodes it depends on, you can perform a depth-first search (DFS) to traverse the dependency graph. Here's how you can implement this: Represent the nodes and their dependencies using a dictionary. Perform a DFS on each node to find the maximum depth of the dependencies. Keep track of the maximum depth encountered dur"See full answer

Mahaboob P. - "int[] sqSorted(int[] nums) { int i = 0, j = nums.length-1; int k = nums.length-1; int[] sqs = new int[nums.length]; while(i n1) { sqs[k--] = n2; j--; } else { sqs[k--] = n1; i++; } } for(int n: sqs) System.out.println(n); return sqs; }"See full answer

Rjj - "I tried solving this question and here is the recorded video for the entire solution - https://youtu.be/G_LIbTp58XA Feel free to comment here or on the video for further discussion."See full answer

Bamboo Y. - "Since the problem asks for a O(logN) solution, I have to assume that the numbers are already sorted, meaning the same number are adjacent to each other, the value of the numbers shouldn't matter, and they expect us to use Binary Search. First, we should analyze the pattern of a regular number array without a single disrupter. Index: 0 1 2 3 4. 5 6. 7. 8. 9 Array:[1, 1, 2, 2, 4, 4, 5, 5, 6, 6] notice the odd indexes are always referencing the second of the reoccurring numbers and t"See full answer

Gopal D. - "(Like a Rummy Game) There are 3 colors of tiles. Each tile has a number 1-9 on it. So the 27-tile set makes a deck. There are 4 decks. (Total = 108 tiles) Tile Colors = {Red, Black, Green} The tiles could be grouped together in patterns Types of patterns : The three tiles are identical (R2, R2, R2) The three tiles are of same color and sequential (R2, R3, R4). Sequence cannot be overlapping (R8, R9, R1 is not a pattern) Help : Tile Notations - R2 denotes Red tile having num 2 A player"See full answer

Yenenash W. - " A higher confidence level leads to a wider interval because we are more certain that the parameter of interest lies within that range. Conversely, a lower confidence level results in a narrower interval, but it also means we are less confident that the interval contains the parameter of interest.""See full answer

Pavani B. - "Request inputs/feedback on my response below with regards to this question and ways to improve and if it looks apt for this question. As a TPM at xxx, I was responsible for delivering a feature that would provide prescription (medications) alternatives to users in the Prescription Shopping Consumer Experience App. The project had a tight deadline as it was something that our account management team promised to a client who wanted it for their employees. My goal was to ensure smooth development"See full answer

Scott S. - " To clarify on what we are trying to solve for and get the definition of a distributed scheduler straight . So A Distributed Scheduler refers to a system that can handle job scheduling and execution in a scalable and fault-tolerant manner by distributing the workload across multiple servers or nodes. we will make some assumptions with the design , as follows: we will assume 10,000 QPS for job submissions and should be able query for jobs status . Job execution should have minimal latency"See full answer

Sarvesh G. - "def calc(expr): ans = eval(expr) return ans your code goes debug your code below print(calc("1 + 1")) `"See full answer

Rishi G. - "Problem Statement: The Fibonacci sequence is defined as F(n) = F(n-1) + F(n-2) with F(0) = 1 and F(1) = 1. The solution is given in the problem statement itself. If the value of n = 0, return 1. If the value of n = 1, return 1. Otherwise, return the sum of data at (n - 1) and (n - 2). Explanation: The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding ones, typically starting with 0 and 1. Java Solution: public static int fib(int n"See full answer

Jeff S. - " function findPrimes(n) { if (n 1; i--) { if (num % i === 0) { notPrimes.add(num); return false; } } return true; } for (let i = 2; i 5 && !notPr"See full answer