Software Engineer Interview Questions

Laura S. - "Clarifying questions: is this a brand new product, or are we improving an existing one? (i.e. are we going to have to migrate an existing codebase or are we starting from scratch?). are we resource-strapped (e.g. # engineers, time)? are there any specific priorities for the product, or should i leave it open-ended? Assume: brand new product, well-resourced, no specific priorities. It seems that there are two sides to this question: 1) technical evaluation of different languages, and 2)"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

Sourabh G. - "Functional Requirement Monitor health, metrics Alert in case of failure/anomaly Visualize the live health Analyse machines on periodic basis Non Functional Should not exert load on machines low latency Highly scalable Logs/Metrics Gathering push - machine gather and send to system and low priority background thread along with batching pull - heart beat check (for offline machines) Processing Real time streaming using Kafka/kinesis + Flink TimeSeries database for stor"See full answer

Anonymous Unicorn - "input_logs = [ f"{senderid} {receiverid} {transaction_count}" "1 2 2", "3 2 42", "2 2 22", "1 1 12", "2 1 1", "2 5 4", "4 2 15" ] input_threshold = 20 exptected_output = [ list of user_ids that made more than 20 transactions sorted by number of transactions in descending order "3", # 42 transactions "2", # 27 transactions (22 + 1 + 4) #"4", # 15 transactions #"1" # 14 transactions (2 + 12 + 1) ] def gettopapi_users(logs, thres"See full answer

Rolemberg J. - "I faced a problem about components re-rendering's and unnecessary requests API's which was causing performance complications in my applicactions. I had a structure that create, edit and delete task, so GET, POST, PUT and DELETE API's methods request was necessary and bring that to compliance without compromissing the performance is hard. I started involving componentes and async functions into the useMemo's and useEffect's to have more control, another improvement was take be careful with global"See full answer

Amparo L. - "The reason I want to work at Doordash is because I’m a really hard worker, I never give up and I’m good at delivering stuff to my teachers at school whenever they have something to drop off to them, I look at the paper and then I read the directions given to me on the ipad to drop it off."See full answer

Ben G. - "Ask Follow up Questions Is this for specific type of user or open? Do we have any past research that has been done? Do we have an idea of company and user goals? Do we have metrics of success? Are there any existing constraints? Why  |  5 min Why is this product or feature important? How does this product benefit customers? What business opportunities does it create? What is our hypothesis? What are our company goals? Who  |  3 min Who are the different types"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

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

Satish murthy D. - "It is very easy to shy away and not make a decision because you might feel the right information is not there to make a decision While dealing with ambiguity it is essential to remain Calm. And Establish what it is you want to achieve by asking a lot of questions to the necessary stakeholders Access the risk, analyze the data at hand, ask the right questions, and be very clear and concise in decision-making"See full answer

Ash I. - "Follow STAR Example: In my most recent role at x, we've been working every quarter to set our quarterly goals or even decide for the year but given in 2022 we've been through a great resignation phase there was a high attrition because of which we have overloaded engineering teams. During last quarterly planning, I started to question with my manager about giving a focus to our engineering, so we complete one thing absolutely fine which is more impactful for our customers and add immediate val"See full answer

Woongshik C. - "Our team were developing a HQ trivia for fitness. So at that moment our focus was developing an eye-catching animation for both iOS and Android. By the way, implementing an animation were hard without the progress of it. So we wanted to see an immediate progress. So I decided to use tools or libraries such as Lottie or Kite. Then it increased the quality and the productivity dramatically. So our designer were happy with it and also our dev team could reuse the code from the designed animation fr"See full answer

Googlepm 1. - "Getting things done : I am extremely proud of walking the extra mile to get things done which I am extremely proud of in my career Ladder transfer : Being on the tactical front, to dream about creating a product from its inception to the delivery and being able to achieve it Gaining trust : Leading without authority is critical and gaining trust in the process of the peers and stakeholders is one thing I am extremely proud of."See full answer

Tiago R. - "function addChildren(root, val, inorder) { const rootInOrderIndex = inorder.indexOf(root.value); const childrenInOrderIndex = inorder.indexOf(val); if (childrenInOrderIndex < rootInOrderIndex) { if (!root.left) { root.left = new TreeNode(val); } else { addChildren(root.left, val, inorder); } } else { if (!root.right) { root.right = new TreeNode(val); } else { addChildren(root.right,"See full answer

Akash A. - "i am a fresher"See full answer

Bravin V. - "I aim to utilize hands-on activities to apply the knowledge I've acquired, safeguard the organization from threat actors, and contribute to its growth."See full answer