Amazon Software Engineer Interview Questions

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

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

Amparo L. - "One time at Lifetime Fitness 2019, One student was throwing a lot of towels in the counter which made things very disorganized, then I made a compromise with my coworker to put 2 towels at a time and then it worked well because it made the workspace very organized."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

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

Scott S. - " The Situation A few months ago, our trading platform started experiencing significant latency issues during peak trading hours. This latency was affecting our ability to process real-time market data and execute trades efficiently, potentially leading to substantial financial losses and missed opportunities. Identifying the Problem The first step was to identify the root cause of the latency. I organized a team meeting with our data engineers, DevOps, and network specialists to gather"See full answer

Avon T. - "Initialize left pointer: Set a left pointer left to 0. Iterate through the array: Iterate through the array from left to right. If the current element is not 0, swap it with the element at the left pointer and increment left. Time complexity: O(n). The loop iterates through the entire array once, making it linear time. Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures. "See full answer

Teddy Y. - "class Node: def init(self, value): self.value = value self.children = [] def inorder_traversal(root): if not root: return [] result = [] n = len(root.children) for i in range(n): result.extend(inorder_traversal(root.children[i])) if i == n // 2: result.append(root.value) if n == 0: result.append(root.value) return result Example usage: root = Node(1) child1 = Node(2) chil"See full answer

Gabriele G. - "DFS with check of an already seen node in the graph would work from collections import deque, defaultdict from typing import List def iscourseloopdfs(idcourse: int, graph: defaultdict[list]) -> bool: stack = deque([(id_course)]) seen_courses = set() while stack: print(stack) curr_course = stack.pop() if currcourse in seencourses: return True seencourses.add(currcourse) for dependency in graph[curr_course]: "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

Aswath B. - "basic drew the diagram"See full answer

Soumya S. - "In 2019, I was given a very important problem to solve. In a team of 3 we had to build a mobility assist device. The customer segment we would go for was something we could decide. The project was very close to me as I had lost someone I loved because of cancer and I saw how reduced mobility was a huge pain point in not being able to do physical activities. My team could only think of elderly people as the main target market. As the Head of Product what I did was: 1) I helped them dive even d"See full answer

Tiago R. - "class Node { constructor(data) { this.data = data; this.left = null; this.right = null; } } function exploreSubtree(node) { let leftHeight = 0; let rightHeight = 0; let maxDiameter = 0; // visit left if (node.left) { const leftVisit = exploreSubtree(node.left); leftHeight = leftVisit.height + 1; maxDiameter = Math.max(maxDiameter, leftVisit.maxDiameter); } // visit right if (node.right) { con"See full answer

Dinesh K. - "i dont know"See full answer

Aditya S. - "I told a story about how our team was focussed on moving a key metric i.e. NPS and to do that we build 3 top requested user feature. Post release the detractors % didn't move even though the detractors request for features shipped went down. Then I connect with users and did some analysis post which we realised that we need to pivot our focus from shipping features to enabling complete workflows for our users i.e. shipping all those feature which are used together in a feature as then only users"See full answer