1704. Determine if String Halves Are Alike

Description of Problem

You are given a string s of even length. Split this string into two halves of equal lengths, and let a be the first half and b be the second half.

Two strings are alike if they have the same number of vowels ('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'). Notice that s contains uppercase and lowercase letters.

Return true if a and b are alike. Otherwise, return false.

Example 1:

Input: s = "book"
Output: true
Explanation: a = "bo" and b = "ok". a has 1 vowel and b has 1 vowel. Therefore, they are alike.

Example 2:

Input: s = "textbook"
Output: false
Explanation: a = "text" and b = "book". a has 1 vowel whereas b has 2. Therefore, they are not alike.
Notice that the vowel o is counted twice.

Constraints:

• 2 <= s.length <= 1000
• s.length is even.
• s consists of uppercase and lowercase letters.

Solution

Code 1 (Rust)

impl Solution {
    pub fn halves_are_alike(s: String) -> bool {
        let k = s.len() / 2;
        let s = s.as_bytes();
        let mut count = 0;
        for i in 0..k {
            let (ch_a, ch_b) = (s[i], s[k+i]);
            match ch_a {
                b'a' | b'A' => count+=1,
                b'e' | b'E' => count+=1,
                b'i' | b'I' => count+=1,
                b'o' | b'O' => count+=1,
                b'u' | b'U' => count+=1,
                _ => {},
            }

            match ch_b {
                b'a' | b'A' => count-=1,
                b'e' | b'E' => count-=1,
                b'i' | b'I' => count-=1,
                b'o' | b'O' => count-=1,
                b'u' | b'U' => count-=1,
                _ => {},
            }

        }

        return count==0;
    }
}

Code 2 (Rust)

Is there any efficient way to iterate characters?