addDigits.cpp

/**
 * Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
 *
 * For example:
 *
 * Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
 *
 * Follow up:
 * Could you do it without any loop/recursion in O(1) runtime?
 *
 * solution approach:
 * Look at the pattern which emerges when we sum the values and eventually reduce it to single digit. By observing carefully, we will see a pattern.
 *
 * This is calculating digital root of a number (https://en.wikipedia.org/wiki/Digital_root)
 * The formula is:
 * dr(n) = 0 if n = 0,
 * 		   9 if n != 0 and  (n % 9 == 0)
 * 		   n % 9
 *
 */

#include <iostream>


int sumDigits( int num ) {
	if ( num == 0 ) {
		return 0;
	} else if ( num % 9 == 0 ) {
		return 9;
	}
	return num % 9;
}

int main() {
	int n;
	std::cout << "Enter a number:";
	std::cin >> n;
	std::cout << "Digital root of " << n << " is :" << sumDigits(n) << std::endl;
	return 0;
}