Recursive Digit Sum HackerRank solution in Java with Explanation

Java solution with explanation for Recursive Digit Sum HackerRank problem

Problem Description :

We define super digit of an integer x using the following rules: 

• Given an integer, we need to find the super digit of the integer.
• If x has only 1 digit, then its super digit is x.
• Otherwise, the super digit of x is equal to the super digit of the sum of the digits of x.
  

For example, the super digit of 9875 will be calculated as:

super_digit(9875)      9 + 8 + 7 + 5 = 29
super_digit(29)          2 + 9 = 11
super_digit(11)          1 + 1 = 2
super_digit(2)            2

Example 1 :

n = 9875
k = 4

The number p is created by concatenating the string n, k times so the initial p = 9875987598759875.

superDigit(p) = superDigit(9875987598759875)
    9+8+7+5+9+8+7+5+9+8+7+5+9+8+7+5 = 116
superDigit(p) = superDigit(116)
    1+1+6 = 8
superDigit(p) = superDigit(8)

All of the digits of p sum to 116. The digits of 116 sum to 8. and 8 is only one digit, so it is the super digit.

Example 2 :

n = 148
k = 3

The number p is created by concatenating the string n, k times so the initial p = 148148148.

superDigit(p) = superDigit(148148148)
    1+4+8+1+4+8+1+4+8 = 39
superDigit(p) = superDigit(39)
    3+9 = 12
superDigit(p) = superDigit(12)
    1+2 = 3

All of the digits of p sum to 39. The digits of 39 sum to 12. The digits of 12 sum to 3. and 3 is only one digit, so it is the super digit.

So here we have to full fill following condition for finding solution :

1. Add n string to k times. 
2. Find sum of all digits.
1. If sum of all digit is 1 number (less than 10), then return.
2. Otherwise again find sum of answered number.

So lets jump on solution and then its explanation :

Solution 1 :

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

    public static int superDigit(String n, int k) {
        char[] c = n.toCharArray();
        long sum = 0;
        for (int i = 0; i < c.length; i++) {
            int a = Integer.parseInt(String.valueOf(c[i]));
            sum = sum + a;
        }
        
        sum = sum * k;
        char[] chars = ("" + sum).toCharArray();
        return getTotalSum(chars);
    }
    
    public static int getTotalSum(char[] array) {
        if(array.length == 1 && Character.getNumericValue(array[0]) <= 9) {
            return  Character.getNumericValue(array[0]);
        }
        int ans = 0;
        for (char c : array) {
            ans = ans + Character.getNumericValue(c);
        }
        array = ("" + ans).toCharArray();
        return getTotalSum(array);
    }
}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        String[] firstMultipleInput = bufferedReader.readLine().replaceAll("\\s+$", "").split(" ");

        String n = firstMultipleInput[0];

        int k = Integer.parseInt(firstMultipleInput[1]);

        int result = Result.superDigit(n, k);

        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Solution explanation :

• First, converting given String n to character array using toCharArray() method.
• Store sum of all character values (numbers) to int sum variable.
• Multiply sum value by given int k value. (As we need to do concatenating initial n to k).
• Convert int value to character array. (Because we need sum of all number characters).

    In getTotalSum() ->

• First check passed character array length, if its length is 1 and contains number that is less then 10 return that. (As this is our answer) This is base condition for recursion call.
• Otherwise traverse through all numbers character, do sum of that and store into int ans variable. (Use getNumericValue() method for get particular number from char array, otherwise we will get character value not our specific number).
  
• Convert int ans variable to character array and call getTotalSum() method recursively.
  

 

Happy learning. Happy coding.

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