# Number of times a number can be replaced by the sum of its digits until it only contains one digit

Count the number of times a number can be replaced by the sum of its digits until it only contains one digit and number can be very large.
Examples:

Input : 10
Output : 1
1 + 0 = 1, so only one times
an number can be replaced by its sum .

Input : 991
Output : 3
9 + 9 + 1 = 19, 1 + 9 = 10, 1 + 0 = 1
hence 3 times the number can be replaced
by its sum.

We have discussed Finding sum of digits of a number until sum becomes single digit
The problem here is just extension of the above previous problem. Here, we just want to count number of times a number can be replaced by its sum until it only contains one digit. As number can be very much large so to avoid overflow, we input the number as string. So, to compute this we take one variable named as temporary_sum in which we repeatedly calculate the sum of digits of string and convert this temporary_sum into string again. This process repeats till the string length becomes 1 . To explain this in a more clear way consider number 991
9 + 9 + 1 = 19, Now 19 is a string
1 + 9 = 10, again 10 is a string
1 + 0 = 1 . again 1 is a string but here string length is 1 so, loop breaks .
The number of sum operations is the final answer .
Below is implementation of this approach .

## C++

 // C++ program to count number of times we // need to add digits to get a single digit. #include using namespace std;   int NumberofTimes(string str) {     // Here the count variable store     // how many times we do sum of     // digits and temporary_sum     // always store the temporary sum     // we get at each iteration .     int temporary_sum = 0, count = 0;       // In this loop we always compute     // the sum of digits in temporary_     // sum variable and convert it     // into string str till its length     // become 1 and increase the count     // in each iteration.     while (str.length() > 1)     {         temporary_sum = 0;           // computing sum of its digits         for (int i = 0; i < str.length(); i++)             temporary_sum += ( str[ i ] - '0' ) ;           // converting temporary_sum into string         // str again .         str = to_string(temporary_sum) ;           // increase the count         count++;     }       return count; }   // Driver program to test the above function int main() {     string s = "991";     cout << NumberofTimes(s);     return 0; }

## Java

 // Java program to count number of times we // need to add digits to get a single digit.   public class GFG {     static int NumberofTimes(String str)     {         // Here the count variable store         // how many times we do sum of         // digits and temporary_sum         // always store the temporary sum         // we get at each iteration .         int temporary_sum = 0, count = 0;                // In this loop we always compute         // the sum of digits in temporary_         // sum variable and convert it         // into string str till its length         // become 1 and increase the count         // in each iteration.         while (str.length() > 1)         {             temporary_sum = 0;                    // computing sum of its digits             for (int i = 0; i < str.length(); i++)                 temporary_sum += ( str.charAt(i) - '0' ) ;                    // converting temporary_sum into string             // str again .             str = temporary_sum + "" ;                    // increase the count             count++;         }                return count;     }           // Driver program to test above functions     public static void main(String[] args)     {          String s = "991";          System.out.println(NumberofTimes(s));     }   } /* This code is contributed by Mr. Somesh Awasthi */

## Python3

 # Python 3 program to count number of times we # need to add digits to get a single digit. def NumberofTimes(s):       # Here the count variable store     # how many times we do sum of     # digits and temporary_sum     # always store the temporary sum     # we get at each iteration .     temporary_sum = 0     count = 0       # In this loop we always compute     # the sum of digits in temporary_     # sum variable and convert it     # into string str till its length     # become 1 and increase the count     # in each iteration.     while (len(s) > 1):               temporary_sum = 0           # computing sum of its digits         for i in range(len(s)):             temporary_sum += (ord(s[ i ]) -                               ord('0'))           # converting temporary_sum into         # string str again .         s = str(temporary_sum)           # increase the count         count += 1       return count   # Driver Code if __name__ == "__main__":           s = "991"     print(NumberofTimes(s))   # This code is contributed by Ita_c

## C#

 // C# program to count number of // times we need to add digits to // get a single digit. using System;   class GFG {           // Function to count number of     // times we need to add digits     // to get a single digit     static int NumberofTimes(String str)     {                   // Here the count variable store         // how many times we do sum of         // digits and temporary_sum         // always store the temporary sum         // we get at each iteration .         int temporary_sum = 0, count = 0;               // In this loop we always compute         // the sum of digits in temporary_         // sum variable and convert it         // into string str till its length         // become 1 and increase the count         // in each iteration.         while (str.Length > 1)         {             temporary_sum = 0;                   // computing sum of its digits             for (int i = 0; i < str.Length; i++)                 temporary_sum += (str[i] - '0');                   // converting temporary_sum             // into string str again .             str = temporary_sum + "" ;                   // increase the count             count++;         }               return count;     }           // Driver code     public static void Main()     {         String s = "991";         Console.Write(NumberofTimes(s));     }   }   // This code is contributed by Nitin Mittal.

## PHP

 1)     {         \$temporary_sum = 0;           // computing sum of its digits         for (\$i = 0; \$i < strlen(\$str); \$i++)             \$temporary_sum += (\$str[ \$i ] - '0');           // converting temporary_sum into         // string str again .         \$str = (string)(\$temporary_sum);           // increase the count         \$count++;     }       return \$count; }   // Driver Code \$s = "991"; echo NumberofTimes(\$s);   // This code is contributed // by Akanksha Rai ?>

## Javascript



Output:

3

Time complexity: O(d2) where d is no of digits in given number n

Auxiliary Space: O(1)
If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.