Digital Root of a given large number using Recursion
Given a large number num in the form of string with length as N, the task is to find its digital root.
The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit.
Examples:
Input: num = 675987890789756545689070986776987
Output: 5
Explanation:
Sum of individual digit of the above number = 212
Sum of individual digit of 212 = 5
So the Digital root is 5Input: num = 876598758938317432685778263
Output: 2
Explanation:
Sum of individual digit of the above number = 155
Sum of individual digit of 155 = 11
Sum of individual digit of 11 = 2
So the Digital root is 2
Approach:
Follow the below steps to solve the problem:
- Find out all the digits of a number.
- Add all the number one by one.
- If the final sum contains more than one digit, Call the recursive function again to make it a single digit.
- The result obtained in the single-digit is the Digital Root of the number.
Below is the implementation of the above approach:
C++
// C++ program to print the digital // root of a given very large number #include <bits/stdc++.h> using namespace std; // Function to convert given // sum into string string convertToString( int sum) { string str = "" ; // Loop to extract digit one by one // from the given sum and concatenate // into the string while (sum) { // Type casting for concatenation str = str + ( char )((sum % 10) + '0' ); sum = sum / 10; } // Return converted string return str; } // Function to get individual digit // sum from string string GetIndividulaDigitSum(string str, int len) { int sum = 0; // Loop to get individual digit sum for ( int i = 0; i < len; i++) { sum = sum + str[i] - '0' ; } // Function call to convert // sum into string return convertToString(sum); } // Function to calculate the digital // root of a very large number int GetDigitalRoot(string str) { // Base condition if (str.length() == 1) { return str[0] - '0' ; } // Function call to get // individual digit sum str = GetIndividulaDigitSum(str, str.length()); // Recursive function to get digital // root of a very large number return GetDigitalRoot(str); } // Driver code int main() { string str = "675987890789756545689070986776987" ; // Function call cout << GetDigitalRoot(str); } |
Java
// Java program to print the digital // root of a given very large number import java.util.*; import java.io.*; class GFG { // Function to convert given // sum into String static String convertToString( int sum) { String str = "" ; // Loop to extract digit one by one // from the given sum and concatenate // into the String while (sum > 0 ) { // Type casting for concatenation str = str + ( char )((sum % 10 ) + '0' ); sum = sum / 10 ; } // Return converted String return str; } // Function to get individual digit // sum from String static String GetIndividulaDigitSum(String str, int len) { int sum = 0 ; // Loop to get individual digit sum for ( int i = 0 ; i < len; i++) { sum = sum + str.charAt(i) - '0' ; } // Function call to convert // sum into String return convertToString(sum); } // Function to calculate the digital // root of a very large number static int GetDigitalRoot(String str) { // Base condition if (str.length() == 1 ) { return str.charAt( 0 ) - '0' ; } // Function call to get // individual digit sum str = GetIndividulaDigitSum(str, str.length()); // Recursive function to get digital // root of a very large number return GetDigitalRoot(str); } // Driver code public static void main(String[] args) { String str = "675987890789756545689070986776987" ; // Function call System.out.print(GetDigitalRoot(str)); } } // This code is contributed by sapnasingh4991 |
Python3
# Python3 program to print the digital # root of a given very large number # Function to convert given # sum into string def convertToString( sum ): str1 = "" # Loop to extract digit one by one # from the given sum and concatenate # into the string while ( sum ): # Type casting for concatenation str1 = str1 + chr (( sum % 10 ) + ord ( '0' )) sum = sum / / 10 # Return converted string return str1 # Function to get individual digit # sum from string def GetIndividulaDigitSum(str1, len1): sum = 0 # Loop to get individual digit sum for i in range (len1): sum = sum + ord (str1[i]) - ord ( '0' ) # Function call to convert # sum into string return convertToString( sum ) # Function to calculate the digital # root of a very large number def GetDigitalRoot(str1): # Base condition if ( len (str1) = = 1 ): return ord (str1[ 0 ]) - ord ( '0' ) # Function call to get # individual digit sum str1 = GetIndividulaDigitSum(str1, len (str1)) # Recursive function to get digital # root of a very large number return GetDigitalRoot(str1) # Driver code if __name__ = = '__main__' : str1 = "675987890789756545689070986776987" # Function call print (GetDigitalRoot(str1)) # This code is contributed by Surendra_Gangwar |
C#
// C# program to print the digital // root of a given very large number using System; class GFG { // Function to convert given // sum into String static String convertToString( int sum) { String str = "" ; // Loop to extract digit one by one // from the given sum and concatenate // into the String while (sum > 0) { // Type casting for concatenation str = str + ( char )((sum % 10) + '0' ); sum = sum / 10; } // Return converted String return str; } // Function to get individual digit // sum from String static String GetIndividulaDigitSum(String str, int len) { int sum = 0; // Loop to get individual digit sum for ( int i = 0; i < len; i++) { sum = sum + str[i] - '0' ; } // Function call to convert // sum into String return convertToString(sum); } // Function to calculate the digital // root of a very large number static int GetDigitalRoot(String str) { // Base condition if (str.Length == 1) { return str[0] - '0' ; } // Function call to get // individual digit sum str = GetIndividulaDigitSum(str, str.Length); // Recursive function to get digital // root of a very large number return GetDigitalRoot(str); } // Driver code public static void Main(String[] args) { String str = "675987890789756545689070986776987" ; // Function call Console.Write(GetDigitalRoot(str)); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript program to print the digital // root of a given very large number // Function to convert given // sum into string function convertToString(sum) { let str = "" ; // Loop to extract digit one by one // from the given sum and concatenate // into the string while (sum > 0) { // Type casting for concatenation str = str + String.fromCharCode((sum % 10) + '0' .charCodeAt()); sum = parseInt(sum / 10, 10); } // Return converted string return str; } // Function to get individual digit // sum from string function GetIndividulaDigitSum(str, len) { let sum = 0; // Loop to get individual digit sum for (let i = 0; i < len; i++) { sum = sum + str[i].charCodeAt() - '0' .charCodeAt(); } // Function call to convert // sum into string return convertToString(sum); } // Function to calculate the digital // root of a very large number function GetDigitalRoot(str) { // Base condition if (str.length == 1) { return (str[0].charCodeAt() - '0' .charCodeAt()); } // Function call to get // individual digit sum str = GetIndividulaDigitSum(str, str.length); // Recursive function to get digital // root of a very large number return GetDigitalRoot(str); } let str = "675987890789756545689070986776987" ; // Function to print final digit document.write(GetDigitalRoot(str)); // This code is contributed by mukesh07. </script> |
5
Time Complexity: O(N), which can be computed as follows:
- O(log N) time to find the sum of digits of num first time, where N is the count of digits initially.
- O(log log N) time to find the sum of digits next time
- O(log log log N) time to find the sum of digits next time
- … and so on.
Auxiliary Space: O(1), since the sum stored in the string will not exceed 3 digits for N ? 105