Monday, January 25, 2016

Buttercola: Twitter: Top k frequent words in a string


Buttercola: Twitter: Top k frequent words in a string

http://www.1point3acres.com/bbs/thread-143017-1-1.html

给一组string "a a b b c c beta beta beta alpha"
n = 2;
要求返回一个List<String> 含有的是出现次数最多的n个, 也就是说如果n=1 list中只有出现次数最多的那个也就是beta, 如果是n=2, list当中出现beta, a b c, 因为a b c都出现两次,算是第二出线多的。. from: 1point3acres.com/bbs 
所以就是n是几,就返回几个数吧,按出现频率来。

开始的思路是, hashmap, + array存出现次数,再arrays.sort(),在获取hashmap当中的key, 印度小哥说不好, 你再想想,我马上说用一个int max 来记录出现次数最多的,找到最多的然后再找第二多的。. 1point3acres.com/bbs
小哥说,不错,这个方法处理数据不大的时候是可以得,接着问,能不能用one pass的形式?
想了一会,疑问句:"PriorityQueue?"  小哥说,嗯,说说吧,就是每个出现单词出现的次数都扔进priorityQueue里面,它自动排序嘛-google 1point3acres
小哥说,那你打代码吧
这里还有一个情况是每一个word之间都是有空格的,两个pointer, start end一个数空格 一个不数空格来获取每一个word。

打完之后,发现小哥说了句:"that looks good for me" 心里搜一个口气。 
最后小哥又让我写了几个Unit test。. Waral 鍗氬鏈夋洿澶氭枃绔�,
我就随便考虑了三种请况。  1. s = null || ""  2. s = "a b b" n =1   3. s = "a b b a b c" n =2. Waral 鍗氬鏈夋洿澶氭枃绔�,
然后用assert(result.get(i从0到N)).equals(目标单词)
小哥说不错。
求人品,要求不多,让我去onsite见识一下场面就可以了。
(准备的豆子问题没有考,也没有自我介绍,上来就码,弄题意弄了将近十分钟,一是信号不算天好,二是印度口音懂的人都懂)

Understand the problem:
这个题目比较麻烦的地方是要求输出具有相同词频的所有string. 比如 n = 2
是要求输出第一大和第二大的所有词。对于这个例子,输出就是
beta (first largest)
a b c (second largest)

所以如果只用 HashMap + PriorityQueue的话不太容易找出来所有前 K 大的数,因为有重复 (e.g. a a b b c c)。并且重复的数不是一起读到PriorityQueue里的。

一个方法是先对所有的输入做词频统计,然后建立倒排索引(inverted index), 同时用 min Heap 来统计 top K 的词频。因为倒排索引可以把所有同样词频的数放到一起,而且在PriorityQueue里面只存词频,因为倒排索引里面已经记录了相应词频的String 是哪些了。


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