leetcode 318. 最大单词长度乘积

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给定一个字符串数组 words，找到 length(word[i]) * length(word[j]) 的最大值，并且这两个单词不含有公共字母。你可以认为每个单词只包含小写字母。如果不存在这样的两个单词，返回 0。

示例 1:

输入: ["abcw","baz","foo","bar","xtfn","abcdef"]
输出: 16 
解释: 这两个单词为 "abcw", "xtfn"。

示例 2:

输入: ["a","ab","abc","d","cd","bcd","abcd"]
输出: 4 
解释: 这两个单词为 "ab", "cd"。

示例 3:

输入: ["a","aa","aaa","aaaa"]
输出: 0 
解释: 不存在这样的两个单词。

  • my solution (slower)

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    class Solution:
        def maxProduct(self, words):
            """
            :type words: List[str]
            :rtype: int
            """
            d = {}
            for w in words:
                d[w] = set(w)
            
            ans = 0
            for i in range(len(words)-1):
                for j in range(i+1, len(words)):
                    if not (d[words[i]] & d[words[j]]):
                        ans = max(ans, len(words[i]) * len(words[j]))
            return ans

  • a faster solution

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class Solution:
    def maxProduct(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        d = {}
        for word in words:
            mask = 0
            for char in set(word):
                mask |= (1 << (ord(char) - 97))
            d[mask] = max(d.get(mask, 0), len(word))
        
        return max([d[x] * d[y] for x in d for y in d if not x & y] or [0])