Software Engineer Interview Questions

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

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

Ugo C. - "function preorderToInorder(preorder) { let inorder = []; let stack = []; let root = preorder[0]; stack.push(root); for (let i = 1; i 0 && stack[stack.length - 1] 0) { root = stack.pop(); inorder.push(r"See full answer

Anonymous Salmon - "Situation Action Result"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

B. T. - "You are given a string S and a number K as input, and your task is to print S to console output considering that, at most, you can print K characters per line. Example: S = "abracadabra sample" K = 11 Output: abracadabra sample Note that this problem requires the interviewee gather extra requirements from the interviewer (e.g. do we care about multiple white spaces? what if the length of a word is greater than K, ...)"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

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

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

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

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

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

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

Hyder A. - "Control plane has few component, it's like the brain of the cluster, the data plane is the work horse, CP has etcd, api-server, DP has kublet and kube-proxy"See full answer