318. 最大单词长度乘积

解题思路：这题的重点在于如何判断两个字符串是否含有相同的字符

解法一：很慢但很好理解。---12% https://leetcode.com/problems/maximum-product-of-word-lengths/discuss/170947/Python-solution-not-fast-but-easy-to-understand

# 找到list中最长的单词，然后其余元素和它进行比较，如果都有元素相同，找下一个最长的单词
# lines.sort(key=lambda x: len(x))

"""
判断两个字符串中是否有相同元素
find() 方法检测字符串中是否包含子字符串 str ，如果指定 beg（开始） 和 end（结束） 范围，
则检查是否包含在指定范围内，如果指定范围内如果包含指定索引值，返回的是索引值在字符串中的起始位置。
如果不包含索引值，返回-1。
str1 = "Runoob example....wow!!!"
str2 = "exam";

print (str1.find(str2))  #7
-----------------------------------------------------------------------------------
info = 'abca'
print(info.find('a'))      # 从下标0开始，查找在字符串里第一个出现的子串，返回结果：0
0
print(info.find('a', 1))   # 从下标1开始，查找在字符串里第一个出现的子串：返回结果3
3
print(info.find('3'))      # 查找不到返回-1
-1

"""
class Solution(object):
    def maxProduct(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        maxlen = 0
        words = sorted(words,key=len,reverse=True) # 记！
        # print(words)
        for i in range(len(words)-1):
            for j in range(i+1,len(words)):
                if len(words[i])*len(words[j])>maxlen:
                    # 如果没有相同的元素
                    # for char in words[j]:
                    #     if char not in words[i]:
                    #         print(char)
                    if not any(char in words[i] for char in words[j]):
                        maxlen = max(maxlen,len(words[i])*len(words[j]))
                        # break
        return maxlen 

解法二：Simple python solution using hashmap (or hashset) ----32%

https://leetcode.com/problems/maximum-product-of-word-lengths/discuss/150099/Simple-python-solution-using-hashmap-(or-hashset) 

# O(n^2+nk) time, where n is a number of words, and k is the length of longest word
# O(n) space
	
def maxProduct(self, words):
    """
    :type words: List[str]
    :rtype: int
    """
    
    # O(k) time
    def get_counts(word): 
        counts_arr = {}
        for char in word:
            if char not in counts_arr:
                counts_arr[char] = 1
            else:
                counts_arr[char] += 1
        return counts_arr
    
    # O(1) time
    def have_common_letters(char_dict1, char_dict2):
        for char in char_dict1:
            if char in char_dict2:
                return True
        return False

   # O(nk) time
    char_counts = {}
    for word in words:
        char_counts[word] = get_counts(word)
      
    # O(n^2) time  
    max_prod_len = 0
    for i in range(len(words)):
        for j in range(i+1, len(words)):
            
            if not have_common_letters(char_counts[words[i]], char_counts[words[j]]):
                max_prod_len = max(max_prod_len, len(words[i])*len(words[j]))
                
    return max_prod_len