Count of digits after concatenation of first N positive integers

Given a positive integer N, the task is to find the total number of digits in the concatenation of the first N positive integers.
Examples:

Input: N = 10 
Output: 11 
Explanation: 
The number formed is 12345678910. 
Hence, the total number of digits = 11
Input: N = 20 
Output: 31 
Explanation:
The number formed is 1234567891011121314151617181920 
Hence, the total number of digits = 31 

Approach: Lets make an observation with the examples. 

• Let N = 13. So, the digits present in all the numbers between 1 to 13 at one’s place are 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3.
• Similarly, the digits at the tens place are 1, 1, 1, 1.
• So, total ones place digits from 1 to 13 are 13(13 – 0).
• Similarly, the total tens place digits are 4(13 – 9).
• Now, lets look at another number to understand the pattern. Let N = 234. So, the digits at units 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.
• Similarly, the digits at tens place are 234 – 9 = 225 because from 1 to 234, only 1 – 9 are the single-digit numbers.
• Lastly, the number of digits at the hundredths place is 234 – 99 = 225 as from 1 to 234, only 1 – 9 are the single-digit numbers and 1 – 99 are the double-digit numbers.
• Therefore, the total number of digits formed when concatenated is 234(234 – 1 + 1) + 225(234 – 10 + 1) + 135(234 – 100 + 1) = 594.
• Hence, the idea is to subtract 0, 9, 99, 999 …. from N to get the number of digits at every place and the summation of all of this is the required answer.

Below is the implementation of the above approach: 

C++

// C++ program to find the number of
// digits after concatenating the
// first N positive integers

#include <iostream>
#include <math.h>
using namespace std;

// Function to find the number of
// digits after concatenating the
// first N positive integers
void numberOfDigits(int N)
{
    int nod = floor(log10(N) + 1);
    int toDecrease
        = (pow(10, nod) - 1) / 9;
    cout << (N + 1) * nod - toDecrease
         << endl;
}

// Driver code
int main()
{
    int N = 13;
    numberOfDigits(N);

    return 0;
}

Java

// Java program to find the number of
// digits after concatenating the
// first N positive integers
class GFG{

// Function to find the number of
// digits after concatenating the
// first N positive integers
static void numberOfDigits(int N)
{
    int nod = (int)Math.floor(Math.log10(N) + 1);
    int toDecrease = (int)(Math.pow(10, nod) - 1) / 9;

    System.out.print((N + 1) * nod - toDecrease);
}

// Driver code
public static void main(String[] args)
{
    int N = 13;
    numberOfDigits(N);
}
}

// This code is contributed by shivanisinghss2110

Python3

# Python3 program to find the number
# of digits after concatenating the
# first N positive integers
from math import log10, floor

# Function to find the number of
# digits after concatenating the
# first N positive integers
def numberOfDigits(N):

    nod = floor(log10(N) + 1);
    toDecrease = (pow(10, nod) - 1) // 9

    print((N + 1) * nod - toDecrease)

# Driver code
if __name__ == '__main__':

    N = 13

    numberOfDigits(N)

# This code is contributed by mohit kumar 29

C#

// C# program to find the number of
// digits after concatenating the
// first N positive integers
using System;
class GFG{

// Function to find the number of
// digits after concatenating the
// first N positive integers
static void numberOfDigits(int N)
{
    int nod = (int)Math.Floor(Math.Log10(N) + 1);
    int toDecrease = (int)(Math.Pow(10, nod) - 1) / 9;

    Console.Write((N + 1) * nod - toDecrease);
}

// Driver code
public static void Main()
{
    int N = 13;
    numberOfDigits(N);
}
}

// This code is contributed by Nidhi_Biet

Javascript

<script>

// JavaScript program to implement
// the above approach

// Function to find the number of
// digits after concatenating the
// first N positive integers
function numberOfDigits(N)
{
    let nod = Math.floor(Math.log10(N) + 1);
    let toDecrease = (Math.pow(10, nod) - 1) / 9;

    document.write((N + 1) * nod - toDecrease);
}

// Driver code

    let N = 13;
    numberOfDigits(N);

 // This code is contributed by sanjoy_62.
</script>

Output:

17

Time Complexity: O(log₁₀N)
Auxiliary Space: O(1)