#### Digital Root

P
```/**
* Author : Suraj Kumar Modi
* https://github.com/skmodi649
*/
/**
* You are given a number n. You need to find the digital root of n.
* DigitalRoot of a number is the recursive sum of its digits until we get a single digit number.
*
* Test Case 1:
* Input:
* n = 1
* Output:  1
* Explanation: Digital root of 1 is 1
*
* Test Case 2:
* Input:
* n = 99999
* Output: 9
* Explanation: Sum of digits of 99999 is 45
* which is not a single digit number, hence
* sum of digit of 45 is 9 which is a single
* digit number.
*/
/**
* Algorithm :
* Step 1 : Define a method digitalRoot(int n)
* Step 2 : Define another method single(int n)
* Step 3 : digitalRoot(int n) method takes output of single(int n) as input
* if(single(int n) <= 9)
* return single(n)
* else
* return digitalRoot(single(n))
* Step 4 : single(int n) calculates the sum of digits of number n recursively
* if(n<=9)
* return n;
* else
* return (n%10) + (n/10)
* Step 5 : In main method simply take n as input and then call digitalRoot(int n) function and
* print the result
*/
package com.thealgorithms.maths;

class DigitalRoot {

public static int digitalRoot(int n) {
if (single(n) <= 9) { // If n is already single digit than simply call single method and
// return the value
return single(n);
} else {
return digitalRoot(single(n));
}
}

// This function is used for finding the sum of the digits of number
public static int single(int n) {
if (n <= 9) { // if n becomes less than 10 than return n
return n;
} else {
return (n % 10) + single(n / 10); // n % 10 for extracting digits one by one
}
} // n / 10 is the number obtained after removing the digit one by one
// The Sum of digits is stored in the Stack memory and then finally returned
}
/**
* Time Complexity: O((Number of Digits)^2) Auxiliary Space Complexity:
* O(Number of Digits) Constraints: 1 <= n <= 10^7
*/
```  