1839. Longest Substring Of All Vowels in Order

Description

A string is considered beautiful if it satisfies the following conditions:

• Each of the 5 English vowels ('a', 'e', 'i', 'o', 'u') must appear at least once in it.
• The letters must be sorted in alphabetical order (i.e. all 'a's before 'e's, all 'e's before 'i's, etc.).

For example, strings "aeiou" and "aaaaaaeiiiioou" are considered beautiful, but "uaeio", "aeoiu", and "aaaeeeooo" are not beautiful.

Given a string word consisting of English vowels, return the length of the longest beautiful substring of word. If no such substring exists, return 0.

A substring is a contiguous sequence of characters in a string.

 

Example 1:

Input: word = "aeiaaioaaaaeiiiiouuuooaauuaeiu"
Output: 13
Explanation: The longest beautiful substring in word is "aaaaeiiiiouuu" of length 13.

Example 2:

Input: word = "aeeeiiiioooauuuaeiou"
Output: 5
Explanation: The longest beautiful substring in word is "aeiou" of length 5.

Example 3:

Input: word = "a"
Output: 0
Explanation: There is no beautiful substring, so return 0.

 

Constraints:

• 1 <= word.length <= 5 * 105
• word consists of characters 'a', 'e', 'i', 'o', and 'u'.

Solution

longest-substring-of-all-vowels-in-order.py

class Solution:
    def longestBeautifulSubstring(self, word: str) -> int:
        if "a" not in word: return 0
        v = ["a", "e", "i", "o", "u"]

        vi = 0
        n = len(word)
        curr = res = 0
        i = j = word.index("a")
        while j < n and word[j] == "a":
            j += 1
        vi += 1

        while i < n and j < n:
            if word[j] != v[vi]:
                vi = 0
                i = j

                while j < n and word[j] != v[vi]:
                    j += 1
                i = j

            if j >= n: return res

            while j < n and word[j] == v[vi]:
                j += 1

            vi += 1
            if vi == 5: 
                res = max(res, j - i)
                vi = 0
                i = j

        return res