Software Engineer Interview Questions

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

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

Ali H. - "SQL databases are relational, NoSQL databases are non-relational. SQL databases use structured query language and have a predefined schema. NoSQL databases have dynamic schemas for unstructured data. SQL databases are vertically scalable, while NoSQL databases are horizontally scalable."See full answer

S R. - "Did pretty ok. Gathered requirements, back of the envelope calculations and drew out the typical system design with details of components. Dug in deep into each box and probed around why it was needed."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

Khushbu R. - "Let’s say the matrix is m x n (i.e., m rows and n columns). Start from the top-right corner of the matrix. Move left if you see a 1. Move down if you see a 0. Keep track of the row index where you last saw the leftmost 1 — that row has the most 1s. public class MaxOnesRow { public static int rowWithMostOnes(int matrix) { int rows = matrix.length; int cols = matrix[0].length; int maxRowIndex = -1; int j = cols - 1; /"See full answer

Shashwat K. - "Binary Search on the array and after than compare the numbers at low and the high pointers whichever is closest is the answer. Because after the binary search low will be pointing to a number which is immediate greater than x and high will be pointing to a number which is immediate lesser than x. int low = 0; int high = n-1; while(low <= high){ int mid = (low + high) / 2; if(x == arr[mid]) return mid; //if x is already present then it will be the closest else if(x < arr[mid]) high"See full answer

Anonymous Jaguar - "Code reviews, automated linting, unit tests, integration tests, validation tests"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

Jagan M. - "This system design question is very small compared to other questions like design instagram, twitter, google drive etc... Since the design involves less components the level of detail we have to go in them were deep. I had to explain how to deal with all the NFR for the distributed cache system. Whether it is a push model or a pull model. Hade to do BOE calculations for the database too."See full answer

Gaurav K. - "I think sliding window will work here and it is the most optimized approach to solve this question."See full answer

Srihitha J. - "my answer: void* memcpy(void* dest, const void* src, size_t n) { unsigned char* uDest = static_cast(dest); const unsigned char* ucSrc = static_cast(src); for(size_t i= 0; i(dest); const unsigned c"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

Anjali M. - "Colleague moved to a different role, so I decided to fill in and did outshine in showing results within the group."See full answer