Software Engineer Interview Questions

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

Stanislav I. - "I want to work at Stripe because Stripe has become the industry standard for many businesses and startups in the world. As a CFO I would be proud to work with a leader on a processing market, improving it position with my skills and experience. I will be happy to be a part of this great Team and learn from them."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

Yag S. - "I studied Exponent's TinyURL system design video. My interviewer was asking many detailed questions on API design, schema, as well as data required to store. I found system design questions are bit high level instead of depth. I think should have detail design of API, schema and some additional flavors."See full answer

GalacticInterviewer - "Approach 1: Use sorting and return the kth largest element from the sorted list. Time complexity: O(nlogn) Approach 2: Use max heap and then select the kth largest element. time complexity: O(n+logn) Approach 3: Quickselect. Time complexity O(n) I explained my interviewer the 3 approaches. He told me to solve in a naive manner. Used Approach 1 had some time left so coded approach 3 also The average time complexity of Quickselect is O(n), making it very efficient for its purpose. However, in"See full answer

Luca D. - "The real discussion was very much similar o what exposed at https://www.tryexponent.com/courses/software-engineering/system-design/design-rate-limiter, but - as I commented the video - the real interviewer wasn't so naive to do not forgive the client identification only because IP. I had to introduce glimpses of https://en.wikipedia.org/wiki/Knowyourcustomer practice, I quoted JWT. I proposed a logical map of id addressing a "deque" of time-stamps of requests, with a threshold for the deque an"See full answer

Dinar M. - "Even more faster and vectorized version, using np.linalg.norm - to avoid loop and np.argpartition to select lowest k. We dont need to sort whole array - we need to be sure that first k elements are lower than the rest. import numpy as np def knn(Xtrain, ytrain, X_new, k): distances = np.linalg.norm(Xtrain - Xnew, axis=1) k_indices = np.argpartition(distances, k)[:k] # O(N) selection instead of O(N log N) sort return int(np.sum(ytrain[kindices]) > k / 2.0) `"See full answer

Jerry O. - "\# Definition for a binary tree node. class TreeNode: def init(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right class Solution: def maxPathSum(self, root: TreeNode) -> int: self.max_sum = float('-inf')"See full answer

Anonymous Jellyfish - "Microservices for resolving diffs, storage S3 for keeping files"See full answer

Rick E. - " from typing import List def getnumberof_islands(binaryMatrix: List[List[int]]) -> int: if not binaryMatrix: return 0 rows = len(binaryMatrix) cols = len(binaryMatrix[0]) islands = 0 for r in range(rows): for c in range(cols): if binaryMatrixr == 1: islands += 1 dfs(binaryMatrix, r, c) return islands def dfs(grid, r, c): if ( r = len(grid) "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

Anonymous Roadrunner - "from typing import List def three_sum(nums: List[int]) -> List[List[int]]: nums.sort() triplets = set() for i in range(len(nums) - 2): firstNum = nums[i] l = i + 1 r = len(nums) - 1 while l 0: r -= 1 elif potentialSum < 0: l += 1 "See full answer

Anonymous Unicorn - "Count items between indices within compartments compartments are delineated by by: '|' items are identified by: '*' input_inventory = "*||||" inputstartidxs = [1, 4, 6] inputendidxs = [9, 5, 8] expected_output = [3, 0, 1] Explanation: "*||||" 0123456789... indices ++ + # within compartments ^ start_idx = 1 ^ end_idx = 9 -- - # within idxs but not within compartments "*||||" 0123456789... indices "See full answer

Mridul J. - "Interviewer wanted to learn more about architecture and or scalability. But I was not sure what the expected answer was"See full answer

Anonymous Wasp - "I was able to answer this question and the follow-up questions as well"See full answer

Anonymous Raccoon - "def find_primes(n): lst=[] for i in range(2,n+1): is_prime=1 for j in range(2,int(i**0.5)+1): if i%j==0: is_prime=0 break if is_prime: lst.append(i) return lst "See full answer

An D. - "Understanding the Basics Choosing Learning Resources Practicing and Applying Knowledge Seeking Help and Staying Updated Leveraging Modern Tools"See full answer