Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/2586.html

2586. Count the Number of Vowel Strings in Range

Description

You are given a 0-indexed array of string words and two integers left and right.

A string is called a vowel string if it starts with a vowel character and ends with a vowel character where vowel characters are 'a', 'e', 'i', 'o', and 'u'.

Return the number of vowel strings words[i] where i belongs to the inclusive range [left, right].

 

Example 1:

Input: words = ["are","amy","u"], left = 0, right = 2
Output: 2
Explanation: 
- "are" is a vowel string because it starts with 'a' and ends with 'e'.
- "amy" is not a vowel string because it does not end with a vowel.
- "u" is a vowel string because it starts with 'u' and ends with 'u'.
The number of vowel strings in the mentioned range is 2.

Example 2:

Input: words = ["hey","aeo","mu","ooo","artro"], left = 1, right = 4
Output: 3
Explanation: 
- "aeo" is a vowel string because it starts with 'a' and ends with 'o'.
- "mu" is not a vowel string because it does not start with a vowel.
- "ooo" is a vowel string because it starts with 'o' and ends with 'o'.
- "artro" is a vowel string because it starts with 'a' and ends with 'o'.
The number of vowel strings in the mentioned range is 3.

 

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length <= 10
• words[i] consists of only lowercase English letters.
• 0 <= left <= right < words.length

Solutions

Solution 1: Simulation

We just need to traverse the string in the interval $[left,.. right]$, and check if it starts and ends with a vowel. If so, the answer plus one.

After the traversal, return the answer.

The time complexity is $O(m)$, and the space complexity is $O(1)$. Where $m = right - left + 1$.

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

• class Solution {
      public int vowelStrings(String[] words, int left, int right) {
          int ans = 0;
          for (int i = left; i <= right; ++i) {
              var w = words[i];
              if (check(w.charAt(0)) && check(w.charAt(w.length() - 1))) {
                  ++ans;
              }
          }
          return ans;
      }
      
      private boolean check(char c) {
          return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
      }
  }
  

• class Solution {
  public:
      int vowelStrings(vector<string>& words, int left, int right) {
          auto check = [](char c) -> bool {
              return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';  
          };
          int ans = 0;
          for (int i = left; i <= right; ++i) {
              auto w = words[i];
              ans += check(w[0]) && check(w[w.size() - 1]);
          }
          return ans;
      }
  };
  

• class Solution:
      def vowelStrings(self, words: List[str], left: int, right: int) -> int:
          return sum(
              w[0] in 'aeiou' and w[-1] in 'aeiou' for w in words[left : right + 1]
          )
  
  

• func vowelStrings(words []string, left int, right int) (ans int) {
  	check := func(c byte) bool {
  		return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'
  	}
  	for _, w := range words[left : right+1] {
  		if check(w[0]) && check(w[len(w)-1]) {
  			ans++
  		}
  	}
  	return
  }
  

• function vowelStrings(words: string[], left: number, right: number): number {
      let ans = 0;
      const check: string[] = ['a', 'e', 'i', 'o', 'u'];
      for (let i = left; i <= right; ++i) {
          var w = words[i];
          if (check.includes(w[0]) && check.includes(w[w.length - 1])) {
              ++ans;
          }
      }
      return ans;
  }
  
  

• impl Solution {
      pub fn vowel_strings(words: Vec<String>, left: i32, right: i32) -> i32 {
          let check = |c: u8| -> bool {
              c == b'a' || c == b'e' || c == b'i' || c == b'o' || c == b'u'
          };
  
          let mut ans = 0;
          for i in left..=right {
              let words_bytes = words[i as usize].as_bytes();
              let first_char = words_bytes[0];
              let last_char = words_bytes[words_bytes.len() - 1];
              if check(first_char) && check(last_char) {
                  ans += 1;
              }
          }
  
          ans
      }
  }
  

All Problems

All Solutions