Software Engineer Interview Questions

Gaurav B. - "class TreeNode(var val: Int, var left: TreeNode? = null, var right: TreeNode? = null) fun isAverageOfDescendants(root: TreeNode?): Boolean { fun helper(node: TreeNode?): Triple { if (node == null) return Triple(0, 0, true) val (leftSum, leftCount, leftValid) = helper(node.left) val (rightSum, rightCount, rightValid) = helper(node.right) val totalSum = leftSum + rightSum val totalCount = leftCount + rightCount // If leaf n"See full answer

Tiago R. - "function isValid(s) { const stack = []; for (let i=0; i < s.length; i++) { const char = s.charAt(i); if (['(', '{', '['].includes(char)) { stack.push(char); } else { const top = stack.pop(); if ((char === ')' && top !== '(') || (char === '}' && top !== '{') || (char === ']' && top !== '[')) { return false; } } } return stack.length === 0"See full answer

Karthik R. - "Good Discussion on the distributed messaging queues (Complex topic with lot of nuances) Liked the mind-map style drawing of requirements and metrics capture Touched on different types of queue styles (point to point, pub-sub, fan-out/fan-in) Storage and WAL usage was interesting Some distributed queue challenges that could be helpful to highlight / expand are: Message guarantee / semantics - Ordering of messages across different servers Replication Master slave architecture or Pe"See full answer

Firdous A. - "Approach (BFS + Horizontal Distance) Assign a horizontal distance (HD) to each node. Root → HD = 0 Left child → HD = parent HD - 1 Right child → HD = parent HD + 1 Do a BFS (level order traversal). If a node with a given HD is seen for the first time, add it to the result. Ignore later nodes with the same HD (because only the top one is visible). After traversal, sort by HD and print nodes left to righ"See full answer

Anonymous Goat - "Batch Packing Problem In Amazon’s massive warehouse inventory, there are different types of products. You are given an array products of size n, where products[i] represents the number of items of product type i. These products need to be packed into batches for shipping. The batch packing must adhere to the following conditions: No two items in the same batch can be of the same product type. The number of items packed in the current batch must be strictly greater than the number pack"See full answer

Rahul J. - "Constraints: 4-direction moves; no mode switching (pick exactly one of {1=bicycle, 2=bike, 3=car, 4=bus} for the full trip). Per-mode search: If a mode’s per-step time/cost are uniform, run BFS on allowed cells. Then totaltime = steps × timeperstep, tie-break by steps × costper_step. If time/cost vary by cell (given matrices), run Dijkstra per mode minimizing (totaltime, totalcost) lexicographically. Maintain the best ⟨time, cost⟩ per cell; relax when the new pair is strictly better. S"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

Babaa - "Function signature for reference: def calculate(servers: List[int], k: int) -> int: ... To resolve this, you can use binary search considering left=0 and right=max(servers) * k so Example: servers=[1,4,5] First server handle 1 request in let's say 1 second, second 4 seconds and last 5 seconds. k=10 So I want to know the minimal time to process 10 requests Get the mid for timeline mid = (left+right)//2 -> mid is 25 Check how many we could process 25//1 = 25 25//4=6 25//5=5 so 25 + 6 +"See full answer

Adam M. - "This is a great question! If you don't mind, I'd love to list 3 strengths and weaknesses then dive a little deeper on one of each of them. I think my three greatest strengths are: Empathy Openness to feedback Drive And three weaknesses I have been working to strengthen are: Sometimes my desire to achieve can put me at risk of overcommitting Communicating status to the right stakeholders at the appropriate time Creating detailed and strategic lists of priorities so I can meet th"See full answer

Kanishvaran P. - "class Solution { public boolean isValid(String s) { // Time Complexity and Space complexity will be O(n) Stack stack=new Stack(); for(char c:s.toCharArray()){ if(c=='('){ stack.push(')'); } else if(c=='{'){ stack.push('}'); } else if(c=='['){ stack.push(']'); } else if(stack.pop()!=c){ return false; } } return stack.isEmpty(); } }"See full answer

Surbhi G. - "Hi Team, where can I see sample answers for this question?"See full answer

Faraz A. - "Scoped out problem constraints - how much data to download, how many computers, and how much time, how you can be detected (decentralized load). Designed a web crawler."See full answer

Namu - "Provided an example from my work history where I asked to move the Go-live date to prevent launch date failure."See full answer

Alok S. - "Functional Rquirements: User should be able to read account balance user should be able to withdraw money from account user should be able to add money to account user should be able to create a new account user should be able to cancel an opened account All of the above for Checkings and Savings account User should be able to pay credit card bill (similar to adding balance to checking and savings account) Non functional Requirements: ACID transactions. consistency > availabi"See full answer

Shashikant A. - "I followed STAR format"See full answer

Sh R. - "i responded using a multi sourced BFS and in place marking, then i checked the final grid to see if any free spots were left unmarked."See full answer

Louie Z. - "DNNs can learn hierarchical features, with each layer learning progressively more abstract features, and generalizes better. SNNs are better for simplier problems involving smaller datasets and if low latency is required."See full answer