Given a string array
Example 1:
words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.Example 1:
Input:["abcw","baz","foo","bar","xtfn","abcdef"]Output:16 Explanation:The two words can be"abcw", "xtfn".
Example 2:
Input:["a","ab","abc","d","cd","bcd","abcd"]Output:4 Explanation:The two words can be"ab", "cd".
Example 3:
Input:["a","aa","aaa","aaaa"]Output:0 Explanation:No such pair of words.
看到这道题之后，我首先试了一下力大砖飞法，但是很不幸，超时了。
于是我观察了一下代码，发现Set这个操作可以在初始化阶段一次性完成。所以加上初始化字典之后就顺利AC了。
但是在我查看Submission的时候，发现了一段特别厉害的Code。仔细拜读了一下之后发现其中的技巧真的非常精妙。
d = {}
for w in words:
mask = 0
for c in set(w):
mask |= (1 << (ord(c) - 97))
d[mask] = max(d.get(mask, 0), len(w))
return max([d[x] * d[y] for x in d for y in d if not x & y] or [0])
这位作者用用Bit-Set位操作的方式避免了使用Python原生Set的比较方式。
真的是一波操作猛如虎，秀的我目瞪口呆。果然还是需要多学习啊。