谷歌之设计题 - 面经题目整理

  1. 设计一个 Map class, 放进去的key value pair 有expiration,也就是ttl。接口是 get(key), put(key, value, ttl)
  2. 设计一个 Timer class,这里有多个不同的版本或变种。一个就是schedule 一个 event,给一定的delay 时间,要求在delay 一定时间后trigger 该event,这里要涉及多线程概念。还有提供你一些辅助函数,比如 set_hw_timer 这种,认为调用该函数就会在一定时间后自动trigger event。大概是不断调用自己函数的写法,另外需要存放一个Heap。比如 狗家店面 新题!
  3. Interface logger 需要 implement两个方法,start和stop,并提供 helper 函数 log 方法。输入是一堆请求。先是对同一个request,当stop被调用,就log这个request的开始时间和结束时间。按格式输出:请求的开始时间和结束时间和请求id。followup是要求按请求开始时间升序。某个请求未结束时不能输出,不考虑多线程。followup需要用heap或priorityqueue来做。
interface Logger { 
	/**
	* When a process starts, it calls 'start' with processId and startTime.
	*/
	void start(String processId, long startTime);
	
	/**
	* When the same process ends, it calls 'end' with processId and endTime.
	*/
	void end(String processId, long endTime);

	/**
	* Prints the logs of this system sorted by the start time of processes in the below format
	* {processId} started at {startTime} and ended at {endTime}
	*/
	void print();
}

Example:

Logger log = new MyLogger();
log.start("1", 100);
log.start("2", 101);  
log.end("2", 102);
log.start("3", 103);
log.end("1", 104);
log.end("3", 105);
log.print();

Output: 
1 started at 100 and ended at 104
2 started at 101 and ended at 102
3 started at 103 and ended at 105
  1. 设计 API Rate Limiter,参考
    https://leetcode.com/discuss/interview-question/124558/Implement-a-Rate-Limiter/

  2. 设计系统来查询 Nearby Restaurants/Friends (a proximity server)

  3. 设计 Web Crawler,参考
    https://leetcode.com/discuss/interview-question/124657/Design-a-distributed-web-crawler-that-will-crawl-all-the-pages-of-wikipedia/

  4. 设计分布式缓存 - 基本就是一致性哈希解释一下

  5. 设计 Load Balancer的数据结构,其实就是 leetcode 原题
    https://leetcode.com/problems/insert-delete-getrandom-o1
    https://leetcode.com/problems/insert-delete-getrandom-o1-duplicates-allowed

  6. 设计 key value store

  7. 设计 Search Autocomplete System
    就是 leetcode 原题 https://leetcode.com/problems/design-search-autocomplete-system

🔎详情点击展开

Design a search autocomplete system for a search engine. Users may input a sentence (at least one word and end with a special character '#' ). For each character they type except ‘#’ , you need to return the top 3 historical hot sentences that have prefix the same as the part of sentence already typed. Here are the specific rules:

  1. The hot degree for a sentence is defined as the number of times a user typed the exactly same sentence before.
  2. The returned top 3 hot sentences should be sorted by hot degree (The first is the hottest one). If several sentences have the same degree of hot, you need to use ASCII-code order (smaller one appears first).
  3. If less than 3 hot sentences exist, then just return as many as you can.
  4. When the input is a special character, it means the sentence ends, and in this case, you need to return an empty list.

Your job is to implement the following functions:

The constructor function:

AutocompleteSystem(String[] sentences, int[] times): This is the constructor. The input is historical data . Sentences is a string array consists of previously typed sentences. Times is the corresponding times a sentence has been typed. Your system should record these historical data.

Now, the user wants to input a new sentence. The following function will provide the next character the user types:

List<String> input(char c): The input c is the next character typed by the user. The character will only be lower-case letters ( 'a' to 'z' ), blank space ( ' ' ) or a special character ( '#' ). Also, the previously typed sentence should be recorded in your system. The output will be the top 3 historical hot sentences that have prefix the same as the part of sentence already typed.

Example:
Operation: AutocompleteSystem([“i love you”, “island”,“ironman”, “i love leetcode”], [5,3,2,2])
The system have already tracked down the following sentences and their corresponding times:
"i love you" : 5 times
"island" : 3 times
"ironman" : 2 times
"i love leetcode" : 2 times
Now, the user begins another search:

Operation: input(‘i’)
Output: [“i love you”, “island”,“i love leetcode”]
Explanation:
There are four sentences that have prefix "i" . Among them, “ironman” and “i love leetcode” have same hot degree. Since ' ' has ASCII code 32 and 'r' has ASCII code 114, “i love leetcode” should be in front of “ironman”. Also we only need to output top 3 hot sentences, so “ironman” will be ignored.

Operation: input(’ ')
Output: [“i love you”,“i love leetcode”]
Explanation:
There are only two sentences that have prefix "i " .

Operation: input(‘a’)
Output: []
Explanation:
There are no sentences that have prefix "i a" .

Operation: input(’#’)
Output: []
Explanation:
The user finished the input, the sentence "i a" should be saved as a historical sentence in system. And the following input will be counted as a new search.

Note:

  1. The input sentence will always start with a letter and end with ‘#’, and only one blank space will exist between two words.
  2. The number of complete sentences that to be searched won’t exceed 100. The length of each sentence including those in the historical data won’t exceed 100.
  3. Please use double-quote instead of single-quote when you write test cases even for a character input.
  4. Please remember to RESET your class variables declared in class AutocompleteSystem, as static/class variables are persisted across multiple test cases . Please see here for more details.
  1. 设计 TicTacToe
  2. 设计 Exam Room, 也可以算是算法题吧,参考
    https://leetcode.com/problems/exam-room/

(持续更新中)

感兴趣的可以看下 谷歌近期面经整理

3 Likes

能不能透露一下Timer这题怎么解?

参考这篇文章

相关文档在这 (搜 Task Scheduler)

类似这个伪代码: set_hw_timer 负责多久后 trigger 这个function,即 run_task

image

相关帖子 狗家店面 新题! - #2 by 999999

谢谢分享!

这题的requirement是:
按照startTime打印,但是能打印的条件是startTime在你之前的request都已经结束了

可以用两个数据结构

// processId -> finish time
Map<String, Long> finish = new HashMap<>();

Queue<Request> q = new PriorityQueue<>();

class Request implements Comparable<Request> {
   long startTime;
   String id;
   
   @Override   
   int compareTo(Request o) {
      return Long.compare(this.startTime, o.startTime);
   }
}

如果assume 我们的input是FIFO的,其实不需要PriorityQueue,直接 LinkedList即可(因为List的顺序就是按照startTime排好序的)

void start(String processId, long startTime) {
     q.add(new Request(processId, startTime));
}

void end(String processId, long endTime) {
    finish.put(processId, endTime);
    Request r = q.peek();
    // 如果不在q的开头,说明有其他没finish的request的startTime在前面
    if (!Objects.equals(r.id, processId)) {
        return;
    }
   
    // 这里碰到有没finish的就停止
    while (r != null && finish.get(r.id) != null) {
        print(r);
       finish.remove(r.id);
        r = q.peek();
    }
}

这题onsite 碰到的

There’re various APIs in an application, and for each API, we’re logging the “start-time” , and the “end-time” for that API execution. We’ve to output all “Slow API” or “Incomplete API” , in the application.

Here’re the definition for slow/incomplete api:
Slow API : execution-time is more than MAX_ALLOWED time. (endTime - startTime > MAX_ALLOWED)
Incomplete API : We do not have any end-time information for this API.

Example Input:
timestamp | API | event-type
1 login start
3 feed start
3 login end
5 trending start
6 profile start
6 feed end
11 trending end

and, MAX_ALLOWED execution-time = 5
NOTE: Input list of logs is sorted in increasing order, based on timestamp values.

Output:
Slow API:

  1. trending

Incomplete API:

  1. profile

** Follow-up:**
Can you write an implementation where, as soon as you find a slow/incomplete API, the program should return the API name?