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2062. Count Vowel Substrings of a String

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

• Java
• C++
• Python
• Go
• TypeScript

• 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';
      }
  };
  

• 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))
  
  

• 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;
  }
  
  

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