1839. Longest Substring Of All Vowels in Order

1839. Longest Substring Of All Vowels in Order (Medium)

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'.

Related Topics:
Two Pointers, String

Solution 1.

// OJ: https://leetcode.com/problems/longest-substring-of-all-vowels-in-order/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int longestBeautifulSubstring(string s) {
        int i = 0, N = s.size(), ans = 0;
        string ch = "aeiou";
        while (i < N) {
            int start = i;
            bool valid = true;
            for (char c : ch) {
                int j = i;
                while (i < N && s[i] == c) ++i;
                valid = i > j;
                if (!valid) break;
            }
            if (valid) ans = max(ans, i - start);
            while (i < N && s[i] != 'a') ++i;
        }
        return ans;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/longest-substring-of-all-vowels-in-order/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int longestBeautifulSubstring(string s) {
        int N = s.size(), cnt = 1, len = 1, ans = 0;
        for (int i = 1; i != N; ++i) {
            if (s[i - 1] == s[i]) ++len;
            else if (s[i - 1] < s[i]) {
                ++len;
                ++cnt;
            } else {
                cnt = 1;
                len = 1;
            }
            if (cnt == 5) ans = max(ans, len);
        }
        return ans;
    }
};