Top Software Engineer Interview Questions

Alemayehu m G. - "I would try to understand their perspective, focus on the issue, find common ground, brainstorm solutions, and agree on a course of action."See full answer

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

Anonymous Unicorn - "python: def justifywords(wordslist, width): result = [] currlinechar_count = 0 curr_words = [] for word in words_list: if curr_words: space_needed = len(word) + 1 # Space needed for the word and a preceding space else: space_needed = len(word) if currlinecharcount + spaceneeded > width: result.append(' '.join(curr_words)) curr_words = [word] currlinechar_count = len("See full answer

Divya R. - "static TreeNode buildTree(int] preorder, int[] inorder) {//[2, 4, 5 if(preorder.length==0 || preorder.length!=inorder.length) { isNull=true; return null; } int parentElem=preorder[0];//2 int index=-1; index=binarySearch(inorder, parentElem);//3 //1 if(index==-1) { isNull=true; return null; } TreeNode root=new TreeNode(parentElem); int preOrder=0; int ind=-1; if(index>0) { int[] leftInOrder=Arrays.copyOfRange(inorder, 0, index); //[4, 2, 5]//[4] ind=binarySearch(preorder, inorder[index"See full answer

Anonymous Kingfisher - "The interviewer focused on the data models and steered the conversation there from the beginning. He had a specific design in mind and I just wasn't hitting it. He also wanted details on how I would implement the logic for splitting users into variants. I would recommend reading the LinkedIn blog on their AB Test system design."See full answer

Vaibhav D. - "Make current as root. 2 while current is not null, if p and q are less than current, go left. If p and q are greater than current, go right. else return current. return null"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

דניאל ר. - "Implemented a recursive function which returns the length of the list so far. when the returned value equals k + 1 , assign current.next = current.next.next. If I made it back to the head return root.next as the new head of the linked list."See full answer

Amparo L. - "I'm currently a School Bus Monitor at First Student, I make sure that the students on the bus are seated down, making sure they are safe during the bus ride. Sometimes depending on the day they may need me to help do a wheelchair so I sometimes help the driver out with making sure the wheelchair is safely secure, hooking up the 4 hooks on the wheelchair, putting the lap belt on the student and also making sure the student feels comfortable during the bus ride. After we finished dropping off our"See full answer

Sandeep Y. - "Functional requirement's: partial search while searching for users, products any keywords in the search. additional keywords in the filter Black listed words in the search. Non functional requirements: low latency, search through 2 Billion records recent search should be cached. Design: high reads, we should have caching enabled over the primary db storages. caching cluster can be added when the search load increases. read ahead. - check in cache (periodic cache refresh), lfu, lru "See full answer

Bulent K. - "I said we could always revert the production issues. Solve it at DEV environment and then publish again. Interviewer gave me an example of it like a database schema change that may not be possible to revert. I said that we always used SQL scripts to upgrade and downgrade the database schema for each production deployment. So it would not be a problem for us. I guess my answer was not accepted."See full answer

Anmol R. - "The example given is wrong. The 2nd test case should have answer 3, as we can get to 6 by using 3 coins of denomination 2."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

Kiran M. - "I started off with developing microservices for the game and then incorporating kafka to make it scalable"See full answer

Tiago R. - "function permute(nums) { if (nums.length <= 1) { return [nums]; } const prevPermutations = permute(nums.slice(0, nums.length-1)); const currentNum = nums[nums.length-1]; const permutations = new Set(); for (let prev of prevPermutations) { for (let i=0; i < prev.length; i++) { permutations.add([...prev.slice(0, i), currentNum, ...prev.slice(i)]); } permutations.add([...prev, currentNum]); } return [...permutations]"See full answer

gdecuri - "Load Balancer to multiple API Servers, pub sub system to publish user events, Game Engine Service, Write through cache and NoSQL DB. HTTP Long Polling to get user status updates"See full answer