Google Software Engineer Interview Questions

Alejandro C. - "1 - Oder list of Kid Position and Sellers Positions (ascending) 2 - Implement a method to check distant 'e' for every kid pos (finding nearest seller and checking if sellerpos - currkid_pos < e, for all kid pos) 3 - Calculate mid from 0 to the 'max post' in between both kids and seller list: (max(max(k) -min(k), max(s) - min(s))) 4 - Perform binary search to find distance 'e' that satisfy step '2'"See full answer

Srini K. - "performance issues and sudden spikes on input requests by scaling techniques and optimization."See full answer

Chetan R. - "Is this in-memory key value store or Disk based as store they are trying to make? Can someone help me with this"See full answer

Graig - "Do any one know how to build trie using IP address?"See full answer

Push S. - "STAR story weaved around the project 0 to 1 execution with all the technical tradeoffs, issues, and success metrics involved."See full answer

David P. - "delete"See full answer

Guilherme M. - "Question: An array of n integers is given, and a positive integer k, where k << n. k indicates that the absolute difference between each element's current index (icurrent) and the index in the sorted array (isorted) is less than k (|icurr - isorted| < k). Sort the given array. The most common solution is with a Heap: def solution(arr, k): min_heap = [] result = [] for i in range(len(arr)) heapq.heappush(min_heap, arr[i]) "See full answer

Jeff S. - " class Node { constructor(data) { this.data = data; this.left = null; this.right = null; } } function diameterOfTree(root) { if (root === null || root.left === null & root.right === null) { return 0; } function countBranch(node, count) { if (node.left === null && node.right === null) { return count; } let left = node.left === null ? 0 : countBranch(node.left, count+1); let right = no"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

Tiago R. - "function serialize(list) { for (let i=0; i 0xFFFF) { throw new Exception(String ${list[i]} is too long!); } const prefix = String.fromCharCode(length); list[i] = ${prefix}${list[i]}; console.log(list[i]) } return list.join(''); } function deserialize(s) { let i=0; const length = s.length; const output = []; while (i < length) { "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

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

Scott S. - " Project Overview: Real-Time Risk Management System Objective The goal was to develop a real-time risk management system capable of processing and analyzing large volumes of trading data to provide near-instantaneous risk assessments. This system was crucial for enabling traders to make informed decisions while managing their exposure to various market risks in real-time. Complexity Factors 1. \\Data Volume and Velocity\\ \\High Throughput:\\ The system needed to ha"See full answer

Santhosh K. - "As an engineering manager, motivation is key to the success of the team. Here are some ways to motivate the team: Set clear goals: Clearly defined goals help team members understand what they're working towards and give them a sense of purpose. Offer growth opportunities: Providing opportunities for professional development and advancement can increase motivation and job satisfaction. Provide recognition and rewards: Recognising and rewarding team members for their hard work and achieve"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

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

Graig - "Yes to peer manager from stackholder team the problem was every time she wanted say something she used to have three way connect with me and my manager"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