# Count total number of digits from 1 to n

Given a number n, count the total number of digits required to write all numbers from 1 to n.

Examples:

```Input : 13
Output : 17
Numbers from 1 to 13 are 1, 2, 3, 4, 5,
6, 7, 8, 9, 10, 11, 12, 13.
So 1 - 9 require 9 digits and 10 - 13 require 8
digits. Hence 9 + 8 = 17 digits are required.

Input : 4
Output : 4
Numbers are 1, 2, 3, 4 . Hence 4 digits are required.
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

To calculate the number of digits, we have to calculate the total number of digits required to write at ones, tens, hundredths …. places of the number . Consider n = 13, so digits at ones place are 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3 and digits at tens place are 1, 1, 1, 1 . So, total ones place digits from 1 to 13 is basically 13 ( 13 – 0 ) and tens place digits is 4 ( 13 – 9 ) . Lets take another example n = 234, so digits at unit place are 1 ( 24 times ), 2 ( 24 times ), 3 ( 24 times ), 4 ( 24 times ), 5 ( 23 times ), 6 ( 23 times ), 7 ( 23 times ), 8 ( 23 times ), 9 ( 23 times ), 0 ( 23 times ) hence 23 * 6 + 24 * 4 = 234 . Digits at tens place are 234 – 9 = 225 as from 1 to 234 only 1 – 9 are single digit numbers . And lastly at hundredths place digits are 234 – 99 = 135 as only 1 – 99 are two digit numbers . Hence, total number of digits we have to write are 234 ( 234 – 1 + 1 ) + 225 ( 234 – 10 + 1 ) + 135 ( 234 – 100 + 1 ) = 594 . So, basically we have to decrease 0, 9, 99, 999 … from n to get the number of digits at ones, tens, hundredths, thousandths … places and sum them to get the required result .

Below is implementation of this approach .

## C/C++

 `// C++ program to count total number of digits ` `// we have to write from 1 to n ` `#include ` `using` `namespace` `std; ` ` `  `int` `totalDigits(``int` `n) ` `{ ` `    ``// number_of_digits store total ` `    ``// digits we have to write ` `    ``int` `number_of_digits = 0; ` ` `  `    ``// In the loop we are decreasing ` `    ``// 0, 9, 99 ... from n till ` `    ``// ( n - i + 1 ) is greater than 0 ` `    ``// and sum them to number_of_digits ` `    ``// to get the required sum ` `    ``for` `(``int` `i = 1; i <= n; i *= 10) ` `        ``number_of_digits += (n - i + 1); ` ` `  `    ``return` `number_of_digits; ` `} ` ` `  `// Driver code to test above function ` `int` `main() ` `{ ` `    ``int` `n = 13; ` `    ``cout << totalDigits(n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to count total number of digits ` `// we have to write from 1 to n ` ` `  `public` `class` `GFG { ` `    ``static` `int` `totalDigits(``int` `n) ` `    ``{ ` `        ``// number_of_digits store total ` `        ``// digits we have to write ` `        ``int` `number_of_digits = ``0``; ` ` `  `        ``// In the loop we are decreasing ` `        ``// 0, 9, 99 ... from n till ` `        ``// ( n - i + 1 ) is greater than 0 ` `        ``// and sum them to number_of_digits ` `        ``// to get the required sum ` `        ``for` `(``int` `i = ``1``; i <= n; i *= ``10``) ` `            ``number_of_digits += (n - i + ``1``); ` ` `  `        ``return` `number_of_digits; ` `    ``} ` ` `  `    ``// Driver Method ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``13``; ` `        ``System.out.println(totalDigits(n)); ` `    ``} ` `} `

## Python 3

 `# Python program to count total number  ` `# of digits we have to write from 1 to n ` ` `  `def` `totalDigits(n): ` ` `  `    ``# number_of_digits store total ` `    ``# digits we have to write ` `    ``number_of_digits ``=` `0``; ` ` `  `    ``# In the loop we are decreasing ` `    ``# 0, 9, 99 ... from n till ` `    ``#( n - i + 1 ) is greater than 0 ` `    ``# and sum them to number_of_digits ` `    ``# to get the required sum ` `    ``for` `i ``in` `range``(``1``, n, ``10``): ` `        ``number_of_digits ``=` `(number_of_digits ``+`  `                                 ``(n ``-` `i ``+` `1``)); ` `         `  `    ``return` `number_of_digits; ` ` `  ` `  `# Driver code ` `n ``=` `13``; ` `s ``=` `totalDigits(n) ``+` `1``; ` `print``(s); ` `     `  `# This code is contributed ` `# by Shivi_Aggarwal  `

## C#

 `// C# program to count total number of ` `// digits we have to write from 1 to n ` `using` `System; ` ` `  `public` `class` `GFG { ` `     `  `    ``static` `int` `totalDigits(``int` `n) ` `    ``{ ` `         `  `        ``// number_of_digits store total ` `        ``// digits we have to write ` `        ``int` `number_of_digits = 0; ` ` `  `        ``// In the loop we are decreasing ` `        ``// 0, 9, 99 ... from n till ` `        ``// ( n - i + 1 ) is greater than 0 ` `        ``// and sum them to number_of_digits ` `        ``// to get the required sum ` `        ``for` `(``int` `i = 1; i <= n; i *= 10) ` `            ``number_of_digits += (n - i + 1); ` ` `  `        ``return` `number_of_digits; ` `    ``} ` ` `  `    ``// Driver Method ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 13; ` `         `  `        ``Console.WriteLine(totalDigits(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

``` 17
```

Time Complexity : O(Logn)

This article is contributed by Surya Priy. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : vt_m, Shivi_Aggarwal

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