LeetCode＃2062. Count vowel substrings in strings

subject :

Substring Is a continuous... In a string （ Non empty ） The character sequence of .

Vowel substring yes only By vowels （'a'、'e'、'i'、'o' and 'u'） A substring composed of , And it must contain All five vowel .

Give you a string word , Count and return word in Number of vowel substrings .

Example 1：

Input ：word = "aeiouu"
Output ：2
explain ： The following is a list word Vowel substring in （ Bold part in italics ）：
- "aeiouu"
- "aeiouu"
Example 2：

Input ：word = "unicornarihan"
Output ：0
explain ：word Does not include 5 Kinds of vowels , So there will be no vowel substring .
Example 3：

Input ：word = "cuaieuouac"
Output ：7
explain ： The following is a list word Vowel substring in （ Bold part in italics ）：
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
Example 4：

Input ：word = "bbaeixoubb"
Output ：0
explain ： All substrings containing all five vowels contain consonants , So there is no vowel substring .
 

Tips ：

1 <= word.length <= 100
word It's only made up of lowercase letters

source ： Power button （LeetCode）
link ： Power button

The judgment substring of this question only needs to meet the conditions required by the question :

Substring Is a continuous... In a string （ Non empty ） The character sequence of .

Vowel substring yes only By vowels （'a'、'e'、'i'、'o' and 'u'） A substring composed of , And it must contain All five vowel .

I used double pointer to solve . The key is to deal with the fast pointer in word Action at the end of .

class Solution:
    def countVowelSubstrings(self, word: str) -> int:
        count=0;store="aeiou"
        pre=0;cur=5
        while len(word[pre:cur])==5:
            now=cur
            while cur<=len(word):
                for i in word[pre:cur]:
                    if i in store:
                        a=True
                        continue
                    else:
                        a=False
                        break
                if a:
                    for i in store:
                        if word[pre:cur].count(i)>0:
                            a=True
                            continue
                        else:
                            a=False
                            break
                if a:
                    count+=1
                cur+=1
            pre+=1
            cur=now+1
        return count