Software Engineer Interview Questions

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

Yomna H. - "Evication Strategy: LFU Access Pattern: Write Around"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

Dhruv M. - "During my internship at Inceptra Analytics, our team was working on improving the monthly reporting process for the operations department. I proposed switching from manual Excel reports to a Power BI dashboard to automate and visualize key metrics. However, some team members were hesitant—they felt the current process, although tedious, was more familiar and controllable. My task was to get buy-in from the team to adopt a more efficient reporting method without causing disruption or resistance."See full answer

Victor O. - " import java.util.*; class Solution { // Time Complexity: O(n^2) // Space Complexity: O(n) public static List> threeSum(int[] nums) { // Ensure that the array is sorted first Arrays.sort(nums); // Create the results list to return List> results = new ArrayList(); // Iterate over the length of nums for (int i = 0; i < nums.length-2; i++) { // We will have the first number 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

Anonymous Wombat - "Was the statement very similar to the leetcode or was it changed and only the main idea remained?"See full answer

Anonymous Roadrunner - "from typing import List def two_sum(nums: List[int], target: int) -> List[int]: prevMap = {} for i, n in enumerate(nums): diff = target - n if diff in prevMap: return [prevMap[diff], i] else: prevMap[n] = i return [] debug your code below print(two_sum([2, 7, 11, 15], 9)) `"See full answer

Anonymous Unicorn - "naive solution: def countprefixpairs(words): n = len(words) count = 0 for i in range(n): for j in range(i + 1, n): if words[i].startswith(words[j]) or words[j].startswith(words[i]): count += 1 return count using tries for when the list of words is very long: from collections import Counter class TrieNode: def init(self): self.children = {} self.count = 0 # To count the number of words ending at this node"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 Jellyfish - "Microservices for resolving diffs, storage S3 for keeping files"See full answer

Anonymous Roadrunner - "class ListNode: def init(self, val=0, next=None): self.val = val self.next = next def has_cycle(head: ListNode) -> bool: slow, fast = head, head while fast and fast.next: slow = slow.next fast = fast.next.next if slow == fast: return True return False debug your code below node1 = ListNode(1) node2 = ListNode(2) node3 = ListNode(3) node4 = ListNode(4) creates a linked list with a cycle: 1 -> 2 -> 3 -> 4"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

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