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 numbervoid 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 codeint main(){ int k = 4, d = 4; printNumberWithDR(k, d); return 0;} |
Java
// Java implementation of the above approachimport java.io.*;class GFG {// Function to find a numberstatic 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 numberdef 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 numberfunction 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 Codevar 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)
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