Software Engineer Interview Questions

Jack F. - "conduct direct user research through methods such as interviews, surveys, and focus groups to gather qualitative insights into user preferences, pain points, and behaviors. Additionally, I analyze quantitative data from sources like user analytics, feedback metrics, and market research to identify trends and patterns. Collaborating closely with stakeholders, including customers, internal teams, and subject matter experts, further enriches the requirements gathering process by incorporating diver"See full answer

Ben G. - "Situation - A time I saw a peer struggling was while I was at Google working in Mountain View. A fellow colleague of mine was consistently not meeting standards in his performance reviews and wasn’t sure why. He was upset because Google was his dream company and he wasn’t sure how to improve based on his feedback. Task - I was tasked to design a based VUI framework for our discovery space on smart displays. This colleague was acting as the VUI lead for the project while I acted as t"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

Roger R. - "Get data from database CDC to Kafka; Write from Kafka to a Bucket; Send the batch file to Visa from time to time;"See full answer

Tiago R. - "function visitChildren(node) { let leftSubtreeHeight = 0; let rightSubtreeHeight = 0; let isChildrenBalanced = true; if (node.left) { const { isBalanced, height } = visitChildren(node.left); isChildrenBalanced = isChildrenBalanced && isBalanced; leftSubtreeHeight += height + 1; } if (isChildrenBalanced && node.right) { const { isBalanced, height } = visitChildren(node.right); isChildrenBalanced = isChildrenBalanced && isBalan"See full answer

Alvaro R. - "bool isValidBST(TreeNode* root, long min = LONGMIN, long max = LONGMAX){ if (root == NULL) return true; if (root->val val >= max) return false; return isValidBST(root->left, min, root->val) && isValidBST(root->right, root->val, max); } `"See full answer

Ramachandra N. - "def mergeTwoListsRecursive(l1, l2): if not l1 or not l2: return l1 or l2 if l1.val < l2.val: l1.next = mergeTwoListsRecursive(l1.next, l2) return l1 else: l2.next = mergeTwoListsRecursive(l1, l2.next) return l2 "See full answer

Ds S. - "Initially I asked clarifying questions like whether the tree can be empty or not and asked the interviewer to explain what is meant by left view and the explanation for the sample inputs. Then I came up with the level order traversal approach where we visit each level in the binary tree at once using a queue and at each level print the value of the first node. Interviewer seemed satisfied with the approach and asked me to code it up. Finally gave the time and space complexity of the solution."See full answer

Jaim - "Whenever I find resistance to my idea, my general approach is to : Validate my idea Identify if it is a political reason or is it a valid engineering tech reason Assuming it is not political but it is resource/difference of opinions that are causing resistance. Then I would try to do a quick POC and show a value proposition for the idea to get the buy-in. One similar situation happened The situation was that stakeholders were promised a feature that would decrease their number of ticket"See full answer

Anonymous Narwhal - "I'm seeking a role in which I can be continually challenged to devise innovative and streamlined solutions for complex issues, particularly with regard to the persistent soaking problem. I'm excited about the opportunity to not only excel in my own performance but also to empower and enable my colleagues to perform at their absolute best. My goal is to wield my influence and magnetic qualities to not only attract but also retain top-tier engineering talent, all in alignment with our shared objec"See full answer

Bhaskar B. - "This could be done using two-pointer approach assuming array is sorted: left and right pointers. We need track two sums (left and right) as we move pointers. For moving pointers we will move left to right by 1 (increment) when right sum is greater. We will move right pointer to left by 1 (decrement) when left sum is greater. at some point we will either get the sum same and that's when we exit from the loop. 0-left will be one array and right-(n-1) will be another array. We are not going to mo"See full answer

Jet 1. - "`#include using namespace std; void printNumbersTillN(int n){ if(n_==0){ return; } printNumbersTillN(n-1); // go to the end -> reach 1 cout>_n; printNumbersTillN(n); }`"See full answer

Vaibhav C. - "To determine if a graph is not a tree, you can check for the following conditions: Presence of cycles: A graph is not a tree if it contains cycles. In a tree, there should be exactly one unique path between any two vertices. If you can find a cycle in the graph, it cannot be a tree. Insufficient number of edges: A tree with N vertices will have exactly N-1 edges. If the graph has fewer or more than N-1 edges, then it is not a tree. Disconnected components: A tree is a connected graph, m"See full answer

Moshe S. - "Law is my passion. Traveling all over the world in 5 years"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