Repeated sum of digits

Given an integer N, recursively sum digits of N until we get a single digit.  The process can be described below

If N < 10    
    digSum(N) = N
Else         
    digSum(N) = Sum(digSum(N))

 

Example 1:

Input:
N = 1234
Output:
1
Explanation:
The sum of 1+2+3+4 = 10, 
digSum(x) == 10 Hence 
ans will be 1+0 = 1

 

 

Example 2:

Input:
N = 9999
Output:
9
Explanation:
Check it yourself.

 

Your Task:

You don't need to read input or print anything. Your task is to complete the function repeatedSumOfDigits() which takes an integer N and returns the repeated sum of digits of N.


class Solution:

    def repeatedSumOfDigits (self,N):

        self.N=N

        s=0

        for i in str(N):

            s+=int(i)

        #return s

        

        if s<10:

            return s

        else:

            s1=0

            for i in str(s):

                s1+=int(i)

            if s1<10:

                return s1

            else:

                s2=0

                for i in str(s1):

                    s2+=int(i)

                return s2

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