- Basic to medium level questions on JavaScript. - React JS based coding round. - Medium DS also based question - System design round which was around API and performance. - A managerial round which is unlike most companies done to eliminate/downgrade candidates
Are stock options a deal breaker ?
System Design round - Airline reservation system
Asked popular Leetcode question based on cache (not LRU).
Find if Linked List has loop, if so, find length of the loop.
Design the recommendations list for amazon webpage.
In Person with 4 people (~45min each): 1. Check if binary tree A is contained in binary tree B. Write code on paper. 2. Derive a formula for calculating the angle between the hour hand and minute hand in a clock. 3. Design a parking lot. Discuss what data structures to use.
Amazon needs a simple service to create, read, update and delete customer addresses. 1. Write the interface for this service. What methods are needed? What parameters would each method require and what data would each method return? 2. Using the following assumptions, describe how you would design your system: You need to store 10 TB of addresses. Your service must support 10,000 transactions per second. Each of your databases can only handle 1TB of data. Each of your application servers can only handle 1,000 TPS Question 1: 1. What is the runtime complexity for the code below? 2. What is the space complexity for the code below? Queue<TreeNode> queue = new LinkedList<BinaryTree.TreeNode>() ; public void myMethod(TreeNode root) { if (root == null) return; queue.clear(); queue.add(root); while(!queue.isEmpty()){ TreeNode node = queue.remove(); if(node.left != null) queue.add(node.left); if(node.right != null) queue.add(node.right); Question 2: 1. What is the runtime complexity for the code below? // Precondition: array[] is sorted int search(int array[], int numberToFind, int min, int max) { while (max >= min) { int index = midpoint(min, max); if(array[index] == numberToFind) return index; else if (array[index] < numberToFind) min = index + 1; else max = index - 1; } // numberToFind was not found System.out.println("Not Found"); }
Find the last unique character in a character string
1. Count the duplicate words in a string
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