Google Software Engineer Interview Questions

David P. - "delete"See full answer

V V. - "Simple file system ( Hash Map , LSM trees), and consistent hashing. I failed to discuss about conflicts ( Versioning)"See full answer

Rahul R. - "I would first call out the two types of Google Doodles that are published. First would be the static doodle which just shows a picture, and users can get more information on it by hovering on it or clicking on it. Second would be the dynamic google doodles which are either short videos or games through which users engage a bit more. My top 3 metrics that covers both would be: Click-through rate and post-click CTR - Although different metrics, I clubbed them since they fall under similar"See full answer

Gabriele G. - "Without using a recursive approach, one can perform a post-order traversal by removing the parent nodes from the stack only if children were visited: def diameterOfTree(root): if root is None: return 0 diameter = 0 stack = deque([[root, False]]) # (node, visited) node_heights = {} while stack: curr_node, visited = stack[-1] if visited: heightleft = nodeheights.get(curr_node.left, 0) heightright = nodehe"See full answer

Jordan Z. - "Maybe we can use this solution: 1, connect all the strings together, and add an integer value ahead each string. 2, use Huffmans algorithm to encode the step 1 result, to make the result size smaller. 3, return the root of Huffmans tree. This solution man be slower than the common serialize method, but it can save a lot of memory, I think, at lease doing serialize is mainly for tranfering data or storing data."See full answer

Matthew K. - " function climbStairs(n) { // 4 iterations of Dynamic Programming solutions: // Step 1: Recursive: // if (n <= 2) return n // return climbStairs(n-1) + climbStairs(n-2) // Step 2: Top-down Memoization // const memo = {0:0, 1:1, 2:2} // function f(x) { // if (x in memo) return memo[x] // memo[x] = f(x-1) + f(x-2) // return memo[x] // } // return f(n) // Step 3: Bottom-up Tabulation // const tab = [0,1,2] // f"See full answer

Batman X. - "def friend_distance(friends, userA, userB): step = 0 total_neighs = set() llen = len(total_neighs) total_neighs.add(userB) while len(total_neighs)!=llen: s = set() step += 1 llen = len(total_neighs) for el in total_neighs: nes = neighbours(friends, userA, el) if userA in nes: return step for p in nes: s.add(p) for el in s: total_neighs.add(el) return -1 def neighbours(A,n1, n2): out = set() for i in range(len(A[n2])): if An2: out.add(i) return out"See full answer

fuzzyicecream14 - "Situation: As a Product Manager at Cisco, I was leading the development of a new highly critical product for enterprise customers. Midway through the project, a key engineering team was reassigned due to an urgent security patch, leaving us understaffed with only six weeks left before a critical customer pilot. Task: I had to ensure the product launched on time without sacrificing key features, despite losing half of our engineering team. The challenge was to"See full answer

Ying T. - "def encode(root): if not root: return [] def dfs(node): if not node: return res.append(node.val) res.append(len(node,children)) for child_node in node.children: dfs(child_node) res = [] dfs(root) return res def decode(arr): if not arr: return None n = len(arr) i = 0 def dfs(val, children_count): if children_count == 0: return Node(val) cur_node = Node(val) cur_node.children = [] for j in range(children_count): nonlocal i i += 2 cur_node.children.append(dfs(arr[i], arr[i"See full answer

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

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

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

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

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

Akash A. - "i am a fresher"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

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

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

Tiago R. - "function addChildren(root, val, inorder) { const rootInOrderIndex = inorder.indexOf(root.value); const childrenInOrderIndex = inorder.indexOf(val); if (childrenInOrderIndex < rootInOrderIndex) { if (!root.left) { root.left = new TreeNode(val); } else { addChildren(root.left, val, inorder); } } else { if (!root.right) { root.right = new TreeNode(val); } else { addChildren(root.right,"See full answer