Software Engineer Interview Questions

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

Santhosh K. - "As an engineering manager, motivation is key to the success of the team. Here are some ways to motivate the team: Set clear goals: Clearly defined goals help team members understand what they're working towards and give them a sense of purpose. Offer growth opportunities: Providing opportunities for professional development and advancement can increase motivation and job satisfaction. Provide recognition and rewards: Recognising and rewarding team members for their hard work and achieve"See full answer

Anonymous Capybara - "using a relational database isn't a good choice for this system! we need more availability here than consistency (CAP theorem)"See full answer

Anonymous Mongoose - "Given a Binary Tree, the task is to find its vertical traversal starting from the leftmost level to the rightmost level. If multiple nodes pass through a vertical line, they should be printed as they appear in the level order traversal of the tree. The idea is to traverse the tree using dfs and maintain a hashmap to store nodes at each horizontal distance (HD) from the root. Starting with an HD of 0 at the root, the HD is decremented for left children and incremented for right children. As we"See full answer

Vidhyadhar V. - "public class CircularBuffer { private T[] buffer; private int head; private int tail; private int size; private final int capacity; public CircularBuffer(int capacity) { this.capacity = capacity; this.buffer = (T[]) new Object[capacity]; this.head = 0; this.tail = 0; this.size = 0; } public void enqueue(T item) { if (isFull()) { throw new IllegalStateException("Buffer is full"); } buf"See full answer

Divya R. - "public static Integer[] findLargest(int[] input, int m) { if(input==null || input.length==0) return null; PriorityQueue minHeap=new PriorityQueue(); for(int i:input) { if(minHeap.size()(int)top){ minHeap.poll(); minHeap.add(i); } } } Integer[] res=minHeap.toArray(new Integer[0]); Arrays.sort(res); return res; }"See full answer

Azeezat R. - "/* You are with your friends in a castle, where there are multiple rooms named after flowers. Some of the rooms contain treasures - we call them the treasure rooms. Each room contains a single instruction that tells you which room to go to next. * instructions1 and treasurerooms_1 * lily* --------- daisy sunflower | | | v v v jasmin --> tulip* violet* ----> rose* --> ^ | ^ ^ | "See full answer

Yag S. - "I studied Exponent's TinyURL system design video. My interviewer was asking many detailed questions on API design, schema, as well as data required to store. I found system design questions are bit high level instead of depth. I think should have detail design of API, schema and some additional flavors."See full answer

Dinar M. - "Even more faster and vectorized version, using np.linalg.norm - to avoid loop and np.argpartition to select lowest k. We dont need to sort whole array - we need to be sure that first k elements are lower than the rest. import numpy as np def knn(Xtrain, ytrain, X_new, k): distances = np.linalg.norm(Xtrain - Xnew, axis=1) k_indices = np.argpartition(distances, k)[:k] # O(N) selection instead of O(N log N) sort return int(np.sum(ytrain[kindices]) > k / 2.0) `"See full answer

Nikhil S. - "I think LRU will be the right option for this because dns cache has ttl associated with it and the combinatuon of LRU cache and timer expiry will be the most effectively used solution. "See full answer

fuzzyicecream14 - "Situation: As a Product Manager at Cisco, I was leading the development of a new highly critical product for enterprise customers. Midway through the project, a key engineering team was reassigned due to an urgent security patch, leaving us understaffed with only six weeks left before a critical customer pilot. Task: I had to ensure the product launched on time without sacrificing key features, despite losing half of our engineering team. The challenge was to"See full answer

Stanislav I. - "I want to work at Stripe because Stripe has become the industry standard for many businesses and startups in the world. As a CFO I would be proud to work with a leader on a processing market, improving it position with my skills and experience. I will be happy to be a part of this great Team and learn from them."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

Anonymous Wombat - "Was the statement very similar to the leetcode or was it changed and only the main idea remained?"See full answer

Luca D. - "The real discussion was very much similar o what exposed at https://www.tryexponent.com/courses/software-engineering/system-design/design-rate-limiter, but - as I commented the video - the real interviewer wasn't so naive to do not forgive the client identification only because IP. I had to introduce glimpses of https://en.wikipedia.org/wiki/Knowyourcustomer practice, I quoted JWT. I proposed a logical map of id addressing a "deque" of time-stamps of requests, with a threshold for the deque an"See full answer

Rick E. - " from typing import List def getnumberof_islands(binaryMatrix: List[List[int]]) -> int: if not binaryMatrix: return 0 rows = len(binaryMatrix) cols = len(binaryMatrix[0]) islands = 0 for r in range(rows): for c in range(cols): if binaryMatrixr == 1: islands += 1 dfs(binaryMatrix, r, c) return islands def dfs(grid, r, c): if ( r = len(grid) "See full answer

GalacticInterviewer - "Approach 1: Use sorting and return the kth largest element from the sorted list. Time complexity: O(nlogn) Approach 2: Use max heap and then select the kth largest element. time complexity: O(n+logn) Approach 3: Quickselect. Time complexity O(n) I explained my interviewer the 3 approaches. He told me to solve in a naive manner. Used Approach 1 had some time left so coded approach 3 also The average time complexity of Quickselect is O(n), making it very efficient for its purpose. However, in"See full answer