// Simple program that demonstrates some recursive solutions to problems
public class Recursion2 {
    
    public static void main(String[] args) {
        System.out.println(repeatDigits(348));
        
        int[] array = {1, 2, 3};
        System.out.println("sum of array: " + sum(array));
    
    } 
    
    // returns the integer obtained by replacing ever digit of n
    // with two of that digit. Example: repeatDigits(348) returns 334488.
    public static int repeatDigits(int n) {
        if (n < 0) {
            return -repeatDigits(-n);
        } else if (n < 10) {
            return 11 * n;
        } else {
            int lastDigit = n % 10;
            int leadingDigits = n / 10;
            return repeatDigits(leadingDigits) * 100
              + repeatDigits(lastDigit);
        }
    }
    
    // returns the sum of all the elements in the array.
    // 0 if the array is empty.
    public static int sum(int[] array) {
        return sum(array, 0);
    }
    
    // returns the sum of all the elements in the array starting 
    // at given index. Returns 0 if index is past the end of the array.
    private static int sum(int[] array, int index) {
        if (array.length >= index) {
            return 0;
        } else {
            // array[0] + sum (array starting at index 1)
            //            array[1] + sum (array starting at index 2) == 3 
            //                       array[2] + sum(array starting at index 3)
            return array[index] + sum(array, index + 1);
        }
    }
}