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.




Naive Recursive Method – 
Naive approach to the above problem is to calculate the length of each number from 1 to n, then calculate the sum of the length of each of them. Recursive implementation of the same is – 

Java

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public class Main {
 
    static int findDigits(int n)
    {
        if (n == 1) {
            return 1;
        }
        // Changing number to String
        String s = String.valueOf(n);
         
        // add length of number to  total_sum
        return s.length() + findDigits(n - 1);
    }
    public static void main(String[] args)
    {
        int n = 13;
        System.out.println(findDigits(n));
    }
}

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Python3

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def findDigits(N):
 
    if N == 1:
        return 1
 
    # Changing number to string
    s = str(N)
 
    # Add length of number to total_sum
    return len(s) + findDigits(N - 1)
 
# Driver Code
 
# Given N
N = 13
 
# Function call
print(findDigits(N))
 
# This code is contributed by vishu2908

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C#

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// C# implementation of
// the above approach
using System;
using System.Collections;
class GFG{
   
static int findDigits(int n)
{
  if (n == 1)
  {
    return 1;
  }
 
  // Changing number to String
  string s = n.ToString();
 
  // add length of number to  total_sum
  return s.Length + findDigits(n - 1);
}
   
// Driver Code
public static void Main(string[] args)
{
  int n = 13;
  Console.Write(findDigits(n));
}
}
 
// This code is contributed by rutvik_56

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Output: 

 17





Iterative Method – (Optimized) 
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 ) . Let’s 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 the implementation of this approach.



C++

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// C++ program to count total number
// of digits we have to write
// from 1 to n
#include <bits/stdc++.h>
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
int main()
{
    int n = 13;
   
    cout << totalDigits(n) << endl;
   
    return 0;
}

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Java

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// 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));
    }
}

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Python3

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# Python3 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

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C#

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// 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.

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PHP

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<?php
// PHP program to count
// total number of digits
// we have to write from
// 1 to n
 
// Function that return
// total number of digits
function 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 = 1; $i <= $n; $i *= 10)
        $number_of_digits += ($n - $i + 1);
 
    return $number_of_digits;
}
 
    // Driver Code
    $n = 13;
    echo totalDigits($n);
     
// This code is contributed by vt_m.
?>

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Output: 

 17





Time Complexity : O(Logn)

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