Google Software Engineer Interview Questions

Bini T. - "I understand this is more focused on ML. However, I have a system question. If users allow us to access their location, or they send location via text box, could we use CDNs for the search without hitting our database? We only query the database when we have zero information on location. Other questions: does embedding always guarantee information on location? Do we discharge the user images after we return a prediction? I heard the feedback that we should keep it for future learning. What would"See full answer

Ven E. - "I saw this question in one of my interviews."See full answer

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

Steve M. - "def split_count(s): return 2**(len(s)-1) `"See full answer

Piyush D. - "Honestly, In addition to having the right expertise and skills required for this job, I am really looking forward to being part of this organisation’s journey. I am really confident that I will really exceed your expectations in achieving both mine and the company’s goals. I have worked with different teams in different capacities, managed a team of engineers successfully from very early part of my career and that has made me a very easy person to work with. I get things done easily. I am passio"See full answer

Tiago R. - "function generateParentheses(n) { if (n < 1) { return []; } if (n === 1) { return ["()"]; } const combinations = new Set(); let previousCombinations = generateParentheses(n-1); for (let prev of previousCombinations) { for (let i=0; i < prev.length; i++) { combinations.add(prev.slice(0, i+1) + "()" + prev.slice(i+1)); } } return [...combinations]; } `"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

Amparo L. - "I was a student worker at Gordon's Food Service, Schaumburg, My tasks were vacuuming onion peels, checking expiration dates, cleaning the break room, cleaning the shelves from the Ailes, Stocking stuff on shelves, sweeping the backroom, mopping, Refilling bottles with cleaning supplies and cleaning the fridge glass."See full answer

Nitin P. - "public static void sortBinaryArray(int[] array) { int len = array.length; int[] res = new int[len]; int r=len-1; for (int value : array) { if(value==1){ res[r]= 1; r--; } } System.out.println(Arrays.toString(res)); } `"See full answer

Chinedu ekene O. - "After more than five years at my previous company, I had contributed significantly to many important projects and helped the team navigate critical transitions, such as our shift from a monolithic to a microservices architecture. However, over time, I realized that my goals and the direction of the company were no longer fully aligned. I felt I had outgrown the position and was ready for new challenges, where I could continue to grow both technically and professionally. Ultimately, I’m excited a"See full answer

TreeOfWisdom - "Should this question be BST, not just BT? Otherwise it would not be possible to reconstruct the tree solely based on the array regardless of its order"See full answer

Debajyoti B. - "As far as i know, when we type a web address in the browser- the broswer connects to the DNS server & finds the address of the website we are looking for after finding it, the browser sends an HTTP request message to the server,with a request to send a copy of the website to the user. all of this information is sent using TCP/IP protocol across your internet"See full answer

Anonymous Hyena - "A load balancer accepts requests from clients (e.g. web browsers on the Internet) and backend services (e.g. a web server). Load balancers are useful for replicating backend services onto multiple machines to meet increased demand. The design of a load balancer should address the following questions: What protocols should be supported? IP allows computers to communicate using packets, similarly to how people send letters. IP packets are addressed using an IP address and port number. "See full answer

Guilherme F. - "A much better solution than the one in the article, below: It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods. shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns. My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N)) class ListNode { constructor(val = 0, next = null) { th"See full answer

Scott S. - "To recount an experience from my last project . We were at the initial stages of designing a new data platform to enhance our trading analytics capabilities. The conflict arose over the selection of a data processing framework. I was advocating for Apache Spark due to its scalability and performance benefits, especially for handling large volumes of derivatives data. Another senior team member preferred a different technology they had more experience with, which they believed would be easier to"See full answer

Tiago R. - "function areSentencesSimilar(sentence1, sentence2, similarPairs) { if (sentence1.length !== sentence2.length) return false; for (let i=0; i (w1 === word1 && !visited.has(w2)) || (w2 === word1 && !visited.has(w1))); if (!edge) { "See full answer