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# LeetCode 1839. Longest Substring Of All Vowels in Order

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 * 10^5`
* `word` consists of characters `'a'`, `'e'`, `'i'`, `'o'`, and `'u'`.

## Solution

```
class Solution {
public:
    int longestBeautifulSubstring(string word) {
        int cnt = 1;
        int len = 1;
        int max_len = 0;
        for (int i = 1; i != word.size(); ++i) {
            if (word[i - 1] == word[i]) {
                ++len;
            } else if (word[i - 1] < word[i]) {
                ++len;
                ++cnt;
            } else {
                cnt = 1;
                len = 1;
            }
            
            if (cnt == 5) {
                max_len = max(max_len, len);
            }
        }
        return max_len;
    }
};
```