Problem
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
Solution (Java)
class Solution {
public int vowelStrings(String[] words, int left, int right) {
int count = 0;
for(int i = left; i <= right; i++) {
if(isVowel(words[i].charAt(0)) && isVowel(words[i].charAt(words[i].length() - 1))) {
count++;
}
}
return count;
}
boolean isVowel(char ch) {
return ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u';
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).