Software Engineer Interview Questions

Shikha S. - "Binary serach on E range"See full answer

Aymen D. - "It would have been more interesting to focus on the system design rather than the Trie DS, Interviewee could have just mentioned the Trie and passed to things more important. Interviewee should have focused on the factors on which he wants to scale the API servers, popularity of the query parts ? region may be ? A hash of many factors ? Caches should have definitely be discussed, Cache eviction policies, Cache invalidation managements... Interviewee should have mentioned which kind of API pro"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

Sankar S. - "We will not always have all the required data to make a decision quickly. We need to work with ambiguity effectively by tying up the available facts with intuition."See full answer

דניאל ר. - "Implemented a recursive function which returns the length of the list so far. when the returned value equals k + 1 , assign current.next = current.next.next. If I made it back to the head return root.next as the new head of the linked list."See full answer

Anonymous Possum - "#inplace reversal without inbuilt functions def reverseString(s): chars = list(s) l, r = 0, len(s)-1 while l < r: chars[l],chars[r] = chars[r],chars[l] l += 1 r -= 1 reversed = "".join(chars) return reversed "See full answer

Yugaank K. - "A recursive backtracking solution in python. def changeSigns(nums: List[int], S: int) -> int: res = [] n = len(nums) def backtrack(index, curr, arr): if curr == S and len(arr) == n: res.append(arr[:]) return if index >= len(nums): return for i in range(index, n): add +ve number arr.append(nums[i]) backtrack(i+1, curr + nums[i], arr) arr.pop() "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

Kofi N. - "const mergeIntervals = (intervals) => { const compare = (a, b) => { if(a[0] b[0]) return 1 else if(a[0] === b[0]) { return a[1] - b[1] } } let current = [] const result = [] const sorted = intervals.sort(compare) for(let i = 0; i = b[0]) current[1] = b[1] els"See full answer

Malay K. - "While answering such a question, it is important to focus on personal growth, learning, and how the experience improved your teamwork. You want to demonstrate self-awareness, humility, and an ability to adapt. I had this experience while working on a cross-functional project that involved collaboration between the engineering and marketing teams. In the interview, I shared a story when I misjudged someone and showed unconscious bias towards due their gender, ethnicity or age. It was embarrassin"See full answer

Pathworks P. - "To answer this, I will focus my efforts on explaining the most common type of API used in most modern software development applications - the REST API. For the purpose of simplicity, I will also keep the topics of Authorization and Authentication out of the mix. In essence, an API is a group of logic that takes in a specific set of inputs and responds with a specific set of outputs. This is analogous to going to a drive-thru and placing an order for a meal. When you give an API a bunch of"See full answer

Khushbu R. - "General Approach (using Max-Heap) Use a max-heap (priority queue) of size k. For each point: Compute the distance to P. Push it into the heap. If heap size > k, remove the farthest point. The heap will contain the k closest points to P. import java.util.*; public class KClosestPoints { static class Point { int x, y; public Point(int x, int y) { this.x = x; this.y = y; } // Euclidean distance squared (no need to take square root) p"See full answer

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

Chris R. - "It was like say we have a library A which has a library B as a dependency and so on, how would we determine in the dependency chain that whether there is a circular depedency?"See full answer