public static int superDigit(String n, int k) {
    // Write your code here
        /*
            Realizes that the super_digit of x is the sum of digits of x moduled by 9.
            Then realizes that value is equal to remain of x when divided by 9.
            However, the string is too long :(
            Sum of digits of that string is under 9*100000(according to the constraints)
            Edge case: super_digit(x) = 0 <=> x = 0
            
            Steps:
            1. Calculating the sum of digits.
            2. Multipling the result with k, then calculating the result % 9
            3. Check if the result is 0 and the sum is not 0 -> then return 9
               Else return that result.
               
            Time complexity: O(L) with L is the length of the string
            Space complexity: O(1)
            
            This approach is still able to be better
        */
        
        int sum = 0; //SUM
        for(int i = 0; i < n.length(); i++){
            sum += n.charAt(i) - '0';
        }
        
        int result = sum;
        result %= 9; //Just to ensure no error with long string and big k
        result *= k; //K times
        result %= 9; //The remain after divide by 9
        
        return (result == 0 && sum != 0) ? 9 : result; 
    }