README_EN.md

2062. Count Vowel Substrings of a String

中文文档

Description

A substring is a contiguous (non-empty) sequence of characters within a string.

A vowel substring is a substring that only consists of vowels ('a', 'e', 'i', 'o', and 'u') and has all five vowels present in it.

Given a string word, return the number of vowel substrings in word.

 

Example 1:

Input: word = "aeiouu"
Output: 2
Explanation: The vowel substrings of word are as follows (underlined):
- "aeiouu"
- "aeiouu"

Example 2:

Input: word = "unicornarihan"
Output: 0
Explanation: Not all 5 vowels are present, so there are no vowel substrings.

Example 3:

Input: word = "cuaieuouac"
Output: 7
Explanation: The vowel substrings of word are as follows (underlined):
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"

 

Constraints:

• 1 <= word.length <= 100
• word consists of lowercase English letters only.

Solutions

Solution 1

class Solution:
    def countVowelSubstrings(self, word: str) -> int:
        n = len(word)
        s = set('aeiou')
        return sum(set(word[i:j]) == s for i in range(n) for j in range(i + 1, n + 1))

class Solution {
    public int countVowelSubstrings(String word) {
        int n = word.length();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            Set<Character> t = new HashSet<>();
            for (int j = i; j < n; ++j) {
                char c = word.charAt(j);
                if (!isVowel(c)) {
                    break;
                }
                t.add(c);
                if (t.size() == 5) {
                    ++ans;
                }
            }
        }
        return ans;
    }

    private boolean isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
}

class Solution {
public:
    int countVowelSubstrings(string word) {
        int ans = 0;
        int n = word.size();
        for (int i = 0; i < n; ++i) {
            unordered_set<char> t;
            for (int j = i; j < n; ++j) {
                char c = word[j];
                if (!isVowel(c)) break;
                t.insert(c);
                ans += t.size() == 5;
            }
        }
        return ans;
    }

    bool isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
};

func countVowelSubstrings(word string) int {
	ans, n := 0, len(word)
	for i := range word {
		t := map[byte]bool{}
		for j := i; j < n; j++ {
			c := word[j]
			if !(c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
				break
			}
			t[c] = true
			if len(t) == 5 {
				ans++
			}
		}
	}
	return ans
}

function countVowelSubstrings(word: string): number {
    let ans = 0;
    const n = word.length;
    for (let i = 0; i < n; ++i) {
        const t = new Set<string>();
        for (let j = i; j < n; ++j) {
            const c = word[j];
            if (!(c === 'a' || c === 'e' || c === 'i' || c === 'o' || c === 'u')) {
                break;
            }
            t.add(c);
            if (t.size === 5) {
                ans++;
            }
        }
    }
    return ans;
}

Solution 2

class Solution:
    def countVowelSubstrings(self, word: str) -> int:
        s = set('aeiou')
        ans, n = 0, len(word)
        for i in range(n):
            t = set()
            for c in word[i:]:
                if c not in s:
                    break
                t.add(c)
                ans += len(t) == 5
        return ans