Print a number containing K digits with digital root D

Given a digital root ‘D’ and number of digits ‘K’. The task is to print a number containing K digits that has its digital root equal to D. Print ‘-1’ if such a number does not exist.

Examples:

Input: D = 4, K = 4
Output: 4000
No. of digits is 4.
Sum of digits is also 4.

Input:  D = 0, K = 1
Output: 0

Approach: A key observation to solving this problem is that appending any number of 0s to a number does not change its digital root. Hence D followed by (K-1) 0’s is a simple solution. 

Special case when D is 0 and K is not 1 does not have a solution since the only number with digital root 0 is 0 itself. 

Below is the implementation of the above approach: 

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find a number
void printNumberWithDR(int k, int d)
{

    // If d is 0 k has to be 1
    if (d == 0 && k != 1)
        cout << "-1";

    else {
        cout << d;
        k--;

        // Print k-1 zeroes
        while (k--)
            cout << "0";
    }
}

// Driver code
int main()
{
    int k = 4, d = 4;

    printNumberWithDR(k, d);

    return 0;
}

Java

// Java implementation of the above approach

import java.io.*;

class GFG {


// Function to find a number
static void printNumberWithDR(int k, int d)
{

    // If d is 0 k has to be 1
    if (d == 0 && k != 1)
        System.out.print( "-1");

    else {
        System.out.print(d);
        k--;

        // Print k-1 zeroes
        while (k-->0)
            System.out.print( "0");
    }
}

// Driver code

    public static void main (String[] args) {
            int k = 4, d = 4;

    printNumberWithDR(k, d);
    }
}


//This code is contributed by inder_verma..

Python3

# Python3 implementation of
# the above approach

# Function to find a number
def printNumberWithDR( k, d ) :

    # If d is 0, k has to be 1
    if d == 0 and k != 1 :
        print(-1, end = "")

    else :
        print(d, end = "")
        k -= 1

        # Print k-1 zeroes
        while k :

            print(0,end = "")
            k -= 1


# Driver code
if __name__ == "__main__" :

    k, d = 4, 4

    # Function call
    printNumberWithDR( k, d )

# This code is contributed by
# ANKITRAI1

C#

// C# implementation of the above approach
using System;

class GFG {

// Function to find a number
static void printNumberWithDR(int k, int d)
{

    // If d is 0 k has to be 1
    if (d == 0 && k != 1)
        Console.Write( "-1");

    else {

        Console.Write(d);
        k--;

        // Print k-1 zeroes
        while (k-->0)
            Console.Write( "0");
    }
}

// Driver code
static public void Main ()
{
    int k = 4, d = 4;

    printNumberWithDR(k, d);
}
}

// This code is contributed by ajit.

PHP

<?php
// PHP implementation of the above approach

// Function to find a number
function printNumberWithDR($k, $d)
{

    // If d is 0 k has to be 1
    if ($d == 0 && $k != 1)
        echo "-1";

    else
    {
        echo $d;
        $k--;

        // Print k-1 zeroes
        while ($k--)
            echo "0";
    }
}

// Driver code
$k = 4;
$d = 4;

printNumberWithDR($k, $d);

// This code is contributed
// by akt_mit
?>

Javascript

<script>

// Javascript implementation of the above approach

// Function to find a number
function printNumberWithDR(k, d)
{

    // If d is 0 k has to be 1
    if (d == 0 && k != 1)
        document.write("-1");

    else
    {
        document.write(d);
        k--;

        // Print k-1 zeroes
        while (k-->0)
            document.write("0");
    }
}

// Driver Code
var k = 4, d = 4;

printNumberWithDR(k, d);

// This code is contributed by Ankita saini

</script>

Output:

4000

Time complexity: O(K)

Auxiliary Space: O(1)