1704. Determine if String Halves Are Alike

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 (Java)

class Solution {
    public boolean halvesAreAlike(String s) {
        if (s.length() < 1) {
            return false;
        }
        return countVowel(0, s.length() / 2, s) == countVowel(s.length() / 2, s.length(), s);
    }

    private int countVowel(int start, int end, String s) {
        int c = 0;
        for (int i = start; i < end; i++) {
            char ch = s.charAt(i);
            if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' || ch == 'A'
                    || ch == 'E' || ch == 'I' || ch == 'O' || ch == 'U') {
                c++;
            }
        }
        return c;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).