In this HackerRank Recursive Digit Sum Interview preparation kit problem you need to Complete the function superDigit that must return the calculated super digit as an integer.


HackerRank Recursive Digit Sum Interview preparation kit solution


Problem solution in Python programming.

def sup_digits(x,k):
    a = digsum(x)
    return sup_digit(str(int(a)*k))

def sup_digit(x):
    if len(x) <= 1:
        return x
    else:
        return sup_digit( digsum(x) )

def digsum(x):
    return str(sum((int(i) for i in list(x))))


n, k = input().split()
print( sup_digits(n, int(k)))


Problem solution in Java Programming.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Solution s = new Solution();
        Scanner sc = new Scanner(System.in);
        
        String str_n = sc.next();
        int k = sc.nextInt();
        
        int pSum = Integer.parseInt(s.supdig(str_n));
        pSum *= k;
                        
        String sup = Integer.toString(s.supdig(pSum));
        
        System.out.println(sup);
    }
    
    String supdig(String n) {
        if(n.length() == 1) return n;
        else {
            int np = 0;
            
            for(int i = 0; i < n.length(); i++) {
                np += Character.getNumericValue( n.charAt(i) );    
            }
            
            return supdig(Integer.toString(np));
        }       
    }
    
    int supdig(int n) {
        if(n / 10 == 0) return n;
        else {
            int r = 0;
            
            while(n > 0) {
                r += n % 10;
                n /= 10;
            }
            
            return supdig(r);
        }
    }
}


Problem solution in C++ programming.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int super_digit( long long k ) {
    if ( k < 10 ) {
        return k;
    }
    
    long long sum = 0;
    while( k > 0 ) {
        sum += k % 10;
        k = k / 10;
    }
    return super_digit( sum );
}


long long sum_initial( string number ){
    long long sum = 0;
    
    for( int i = 0; i < number.size(); i++ ) {
        sum += number[i] - '0';
    }
    
    return sum;
}


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */ 
    string  n;
    int k;
    cin >> n >> k;
    long long repeated = sum_initial(n) * k;
    
    long long result = super_digit(repeated);
    
    cout << result << "\n";
    
    
    
    
    return 0;
}


Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    long long sd=0,h;
    int k;
    int c;
    do{
        c=getchar();
        if(c != ' ')
            sd += c -'0'; 
    }while(c != ' ');
    scanf("%d",&k);
    sd *= k;
    
    while(sd > 10){
        h=0;
        while(sd > 0){
            h+= sd %10;
            sd = sd /10;
        }
        sd =h;
    }
    printf("%lld\n",sd);
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}


Problem solution in JavaScript programming.

const N = require('bignumber.js');

let $ = '';

const r = n => n < 10 ? n : f(new N(n).toFixed());

const f = (n, k = 1) => {
  let i = 0;

  for (const c of n) i += +c;

  return r(i * k);
};

process.stdin.on('data', d => $ += d).on('end', () => {
  const [ n, k ] = $.trim().split(/\s+/);
  const ans = f(new N(n).toFixed(), +k);

  console.log(ans);
});