A string is considered beautiful if it satisfies the following conditions:
'a', 'e', 'i', 'o', 'u') must appear at least once in it.'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 * 105word consists of characters 'a', 'e', 'i', 'o', and 'u'.struct Solution;
impl Solution {
fn longest_beautiful_substring(word: String) -> i32 {
let word: Vec<char> = word.chars().collect();
let vowels: Vec<char> = vec!['a', 'e', 'i', 'o', 'u'];
let mut start = 0;
let n = word.len();
let mut res = 0;
'outer: while start < n {
while start < n && word[start] != 'a' {
start += 1;
}
let mut end = start;
for i in 0..5 {
if !(end < n && word[end] == vowels[i]) {
start = end;
continue 'outer;
}
while end < n && word[end] == vowels[i] {
end += 1;
}
}
res = res.max(end - start);
start = end;
}
res as i32
}
}
#[test]
fn test() {
let word = "aeiaaioaaaaeiiiiouuuooaauuaeiu".to_string();
let res = 13;
assert_eq!(Solution::longest_beautiful_substring(word), res);
let word = "aeeeiiiioooauuuaeiou".to_string();
let res = 5;
assert_eq!(Solution::longest_beautiful_substring(word), res);
let word = "a".to_string();
let res = 0;
assert_eq!(Solution::longest_beautiful_substring(word), res);
}