Count total number of digits from 1 to n
Last Updated :
03 May, 2023
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 –
C++
#include <bits/stdc++.h>
using namespace std;
int findDigits(int n)
{
if (n == 1)
{
return 1;
}
string s = to_string(n);
return s.length() + findDigits(n - 1);
}
int main()
{
int n = 13;
cout << findDigits(n) << endl;
return 0;
}
|
Java
public class Main {
static int findDigits(int n)
{
if (n == 1) {
return 1;
}
String s = String.valueOf(n);
return s.length() + findDigits(n - 1);
}
public static void main(String[] args)
{
int n = 13;
System.out.println(findDigits(n));
}
}
|
Python3
def findDigits(N):
if N == 1:
return 1
s = str(N)
return len(s) + findDigits(N - 1)
N = 13
print(findDigits(N))
|
C#
using System;
using System.Collections;
class GFG{
static int findDigits(int n)
{
if (n == 1)
{
return 1;
}
string s = n.ToString();
return s.Length + findDigits(n - 1);
}
public static void Main(string[] args)
{
int n = 13;
Console.Write(findDigits(n));
}
}
|
Javascript
<script>
function findDigits(n)
{
if (n == 1)
{
return 1;
}
let s = n.toString();
return (s.length + findDigits(n - 1));
}
let n = 13;
document.write( findDigits(n) + "<br>");
</script>
|
Output:
17
Time Complexity: O(n log n)
Auxiliary Space: O(n log n)
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++
#include <bits/stdc++.h>
using namespace std;
int totalDigits(int n)
{
int number_of_digits = 0;
for(int i = 1; i <= n; i *= 10)
number_of_digits += (n - i + 1);
return number_of_digits;
}
int main()
{
int n = 13;
cout << totalDigits(n) << endl;
return 0;
}
|
Java
public class GFG {
static int totalDigits(int n)
{
int number_of_digits = 0;
for (int i = 1; i <= n; i *= 10)
number_of_digits += (n - i + 1);
return number_of_digits;
}
public static void main(String[] args)
{
int n = 13;
System.out.println(totalDigits(n));
}
}
|
Python3
def totalDigits(n):
number_of_digits = 0;
for i in range(1, n, 10):
number_of_digits = (number_of_digits +
(n - i + 1));
return number_of_digits;
n = 13;
s = totalDigits(n) + 1;
print(s);
|
C#
using System;
public class GFG {
static int totalDigits(int n)
{
int number_of_digits = 0;
for (int i = 1; i <= n; i *= 10)
number_of_digits += (n - i + 1);
return number_of_digits;
}
public static void Main()
{
int n = 13;
Console.WriteLine(totalDigits(n));
}
}
|
PHP
<?php
function totalDigits($n)
{
$number_of_digits = 0;
for ($i = 1; $i <= $n; $i *= 10)
$number_of_digits += ($n - $i + 1);
return $number_of_digits;
}
$n = 13;
echo totalDigits($n);
?>
|
Javascript
<script>
function totalDigits(n)
{
var number_of_digits = 0;
for(var i = 1; i <= n; i *= 10)
number_of_digits += (n - i + 1);
return number_of_digits;
}
var n = 13;
document.write(totalDigits(n));
</script>
|
Output:
17
Time Complexity : O(Logn)