Reverse String II

Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.

If there are fewer than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and leave the other as original.

LeetCode Problem - 541

class Solution {
    public String reverseStr(String s, int k) {
        // Create a StringBuilder to build the result string
        StringBuilder sb = new StringBuilder();

        // Case 1: If the string length is less than or equal to 'k', reverse the entire string
        if(s.length() <= k){
            sb.append(s);
            return sb.reverse().toString();
        } 
        // Case 2: If the string length is greater than 'k' but less than or equal to '2*k'
        // Reverse the first 'k' characters, and keep the rest as is
        else if (s.length() > k && s.length() <= 2 * k) {
            sb.append(s.substring(0, k));  // Reverse the first 'k' characters
            return (sb.reverse().toString()) + (s.substring(k, s.length()));  // Concatenate the rest of the string
        } 
        // Case 3: If the string length is greater than '2*k'
        // Reverse the first 'k' characters of every 2*k chunk and keep the next 'k' characters as is
        else if (s.length() > 2 * k) {
            for(int i = 0; i < s.length(); i += 2 * k) {
                // Reverse the first 'k' characters in the current 2*k chunk
                String part1 = s.substring(i, Math.min(i + k, s.length()));
                sb.append(new StringBuilder(part1).reverse());

                // Append the next 'k' characters as is (if they exist)
                if(i + k < s.length()) {
                    String part2 = s.substring(i + k, Math.min(i + 2 * k, s.length()));
                    sb.append(part2);
                }
            }
        }
        // Return the final constructed string
        return sb.toString();
    }
}