1704. Determine if String Halves Are Alike(Solution || Leetcode easy || Java)

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:

class Solution {
    public boolean halvesAreAlike(String s) {
        //Convert all uppercase letters to lowercase
       s = s.toLowerCase();
        
        //initialise variables to count the number of vowels o both side
        int count1 = 0;
        int count2 = 0;
        
        //get the middle index
        int n = (s.length())/2;
        
        //run loop for searching vowels
        for(int i = 0; i < n; i++){
            if(s.charAt(i) == 'a' || s.charAt(i) == 'e' || 
               s.charAt(i) == 'i' ||   s.charAt(i) == 'o' || 
               s.charAt(i) == 'u'){
                count1++;
            }
            
            if(s.charAt(n+i) == 'a' || s.charAt(n+i) == 'e' ||
               s.charAt(n+i) == 'i' ||   s.charAt(n+i) == 'o' ||
               s.charAt(n+i) == 'u'){
                count2++;
            }
        }
        
        return count1 == count2;
    }
}

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