Did not ask technical questions, but did ask about my background.
Technical Support Analyst Iii Interview Questions
5,704 technical support analyst iii interview questions shared by candidates
Design the recommendations list for amazon webpage.
Given an array of strings made up of 1s and 0s, formulate an algorithm that determines which strings are sub-strings of others in the input array
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.
Do you know what RAID is?
What would you do if you had 2 calls come in at the same time with equal priority?
Me perguntaram qual foi meu ultimo projeto.
Describe your strengths in data centers.
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"); }
How to handle certain situations within the workplace.
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