# Accepted C++ O(1)-time O(1)-space 1-Line Solution with Detail Explanations

• The problem, widely known as digit root problem, has a congruence formula:

``````https://en.wikipedia.org/wiki/Digital_root#Congruence_formula
``````

For base b (decimal case b = 10), the digit root of an integer is:

• dr(n) = 0 if n == 0
• dr(n) = (b-1) if n != 0 and n % (b-1) == 0
• dr(n) = n mod (b-1) if n % (b-1) != 0

or

• dr(n) = 1 + (n - 1) % 9

Note here, when n = 0, since (n - 1) % 9 = -1, the return value is zero (correct).

From the formula, we can find that the result of this problem is immanently periodic, with period (b-1).

Output sequence for decimals (b = 10):

~input: 0 1 2 3 4 ...
output: 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 ....

Henceforth, we can write the following code, whose time and space complexities are both O(1).

``````class Solution {
public:
return 1 + (num - 1) % 9;
}
};
``````

• It's very smart!

• great explanation! thanks

• I think -1 % 9 = 8. Because the result of mod should be between 0 to 8.

• This post is deleted!

• The question said there is no loop or no recursion.

• It seems that different languages have different implementation for the mod operator when dealing with negative number.

• At least you want to put a return before your recursion call for the addDigits.

• And why would people provide their stupid answer without checking other posts?

• This post is deleted!

• It`s really amazing!

• Don't know why this kind of pure math questions keep appearing in a coding interview... if you don't know what is digital root it will take you quite a while to figure it out, it's definitely not "easy".

• @flando I`m agree with you!

• nice method.

• @zhiqing_xiao ok,well,we are the same,this is my java code:

``````public class Solution {
if(num==0)return 0;
return num%9==0?9:num%9;
}
}
``````

• @zhiqing_xiao Great！

• nice explanation !

• pretty cool!

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