- 45 min
- 准备工作 以前linked，sync async
-
Princple 讲故事（BQ）
KEYWORD: customer obsession，ownership， simplify，Bias for Action，think bigFreelance
业务逻辑：research solution, task partition, tracking process, deployment
VS co-worker: Implement actual HTTP request on angular VS request, express in long term
VS PM 时间估计：insight of task complexity，refactorize previous components to organize them, visualize UI and logic.实习：
业务逻辑challenge：
A typical story is that:
design of restock request 1 week- 1 month
based on customer's factory need, excel, clicking 2 - 3 times, detail editing
Proposing idea, export grouped by different feature or sorted based on amount of items in long termGeneral 技术challenge：大环境 community，小细节stackoverflow， 熟练practice, play with it
国内：
Contribution: go depth refine mechanism for usibility: error responsive and DB connection pool propose a new project, invent tools for monitor, automata script for deployment, immgration long term save much time
- java background
- encapsulate inheritance polymorphism
- 直接用AWS 的example user order transcation
2 代码 实现细节， static return package
3 QA 准备什么问题
时间：2018-1-10
职位：aws 美国大叔
reverse words 复杂度
要求尽可能time／space efficient
idea 及test case
时间：2018-2-7
职位：内推的亚麻Austin的AWS Marketplace组 SDE
1. ArrayList和LinkedList的比较
2. HashMap的基本操作和时间复杂度
这两个除了理论上的复杂度之外，最好说一下实际中的复杂度，比如hashcode设计得不好时对复杂度的影响
3. print out a binary tree in level order
时间：2018-2-11
职位：AWS phone interview，老中
1道是蠡口 嗣酒
一道是近期面经原题（随便一搜AWS就有）
2017-12-12
设计一个停车场，后面不停的加了一些需求，例如两个入口，两个出口怎么设计， 怎么handle两辆车同时进入的 或同时出来的情况， 用个lock， 把数量lock住，一个一个更新
亚麻 电面
时间：2018-1-6
职位：
要求实现一个关注topic的功能
public interface iSubscriber {
// Subscribes a user to a given topic. visit 1point3acres.com for more.
public void subscribe(Integer userId, String topic);
// Returns all the users who have subscribed to a given topic so far. From 1point 3acres bbs
public List<Integer> getSubscribers(String topic);
}
public class Demo implements iSubscriber{
List<Integer> ids;
HashMap<String, List<Integer>> data;
Demo(){
ids = new ArrayList<Integer>();-google 1point3acres
data = new HashMap<String, List<Integer>>();
}
public void subscribe(Integer userId, String topic){-google 1point3acres
if(topic.isEmpty()){
return;
}
// if is a topic
List<Integer> temp;
if(data.get(topic)!=null){
temp = new ArrayList<Integer>();
}else{
temp = data.get(topic);
}
// if the user is already subscribed
if(!temp.contains(userId)){
temp.add(userId);
}
data.put(topic, temp);
}
public List<Integer> getSubscribers(String topic){
return data.get(topic);
}
}
demo.subscribe(123, "abc")
demo.subscribe(123, "abc")
demo.getSubscribes // verify that it only has 1 userId "123"
demo.subscribe(123, "");
//demo.subscribe(0, "aaaa");. from: 1point3acres.com/bbs
demo.subscribe("1.1", "aaaa");
demo.subscribe(0, 123);
// SQL injection part
demo.subscribe(0, "'; select * from any_table;");
// XSS part
demo.subscribe(0, "'<src=javascript> alert(''); </src>");
Follow up: 如果说topic有几个sub topic，你怎么实现
"Basketball"
"Basketball.Bulls"
"Basketball.Warriors"
"Basketball.Bucks"
"Basketball.Bucks.Middleton"
"Basketball.Bucks.Delly"
1. 输入一个二维矩阵，由0和1组成，每个row开始总是连续的0，可能在某一个index后面又突然全变成了1，输出拥有最多1的row的index。
例如输入为：
[[0, 0, 0],
[0, 1, 1],
[0, 0, 1]]
则输出为：
1
Follow up是输出所有的1最多的row的index，也就是list of index of row
2. OOD，设计一个停车场，写了class的框架和重要methods
3. 问问题
1. Return true if the string is a palindrome or an anagram of palindrome. False otherwise.
anagram的意思是字母的任意组合， "aba" == "aab" == "baa"。
例子："aba" => true, "aab" => true, "abc" => false
2. Tranform string a to string b with minimum edits/costs.
只能进行add和remove
例子: "aba" -> "abc" => 2(remove a, add c).
亚麻电面
1. 利口 衣领散 解释面向对象里的多态和继承
2. 利口 衣舞衣
举个依据客户反馈改进产品的例子.
2017-6-22
螺旋矩阵
2017-6-28
wordladder2
tech：LC 而思，mysql left join 和 join 区别
BQ: 介绍自己， 介绍相关经历
时间很短，白人小哥人很好
2017-5-26
面试官：一个在五年工作的senior，考了Least Frequent Used, 没有写对。
然后问了cs基础概念：interface vs abstract class, 实际code中用过什么desig pattern， unix command find vs grep
