# 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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// 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 */`

## Python 3

 `# 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 ` `?> `

Output:

```3
```

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