# 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 $m = right - left + 1$.

• 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) {
const w = words[i];
if (check.includes(w[0]) && check.includes(w.at(-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 w = words[i as usize].as_bytes();
if check(w[0]) && check(w[w.len() - 1]) {
ans += 1;
}
}

ans
}
}