318. Maximum Product of Word Lengths

Posted 2020-06-11

好久没leetcode一下了，以后每天１题，坚持一下～

https://leetcode.com/problems/maximum-product-of-word-lengths/

Total Accepted: 9706 Total Submissions: 24927 Difficulty: Medium
Given a string array 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:
Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
Return 16
The two words can be "abcw", "xtfn".

Example 2:
Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return 4
The two words can be "ab", "cd".

 

sol: 将每个单词对应到一个数。为增加效率，长度也可提前算好。

class Solution(object):
    def maxProduct(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        n = len(words)
        word_map,len_map = [],[]
        for word in words:
            word_map += sum(1<<(ord(x)-ord(‘a‘)) for x in set(word)),
            len_map += len(word),
        
        ans = 0
        for i in range(n):
            for j in range(i+1,n):
                if word_map[i] & word_map[j] == 0:
                    ans = max( len_map[i]*len_map[j], ans )
        
        return ans