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 –
#include <bits/stdc++.h> using namespace std;
int findDigits( int n)
{ if (n == 1)
{
return 1;
}
// Changing number to String
string s = to_string(n);
// Add length of number to total_sum
return s.length() + findDigits(n - 1);
} // Driver code int main()
{ int n = 13;
cout << findDigits(n) << endl;
return 0;
} // This code is contributed by divyeshrabadiya07 |
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));
}
} |
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 |
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 |
<script> function findDigits(n)
{ if (n == 1)
{
return 1;
}
// Changing number to String
let s = n.toString();
// Add length of number to total_sum
return (s.length + findDigits(n - 1));
} // Driver code let n = 13;
document.write( findDigits(n) + "<br>" );
//This code is contributed by Mayank Tyagi </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++ 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;
} |
// 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));
}
} |
# 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 |
// 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 // 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. ?> |
<script> // Javascript program to count total number // of digits we have to write // from 1 to n function totalDigits(n)
{ // number_of_digits store total
// digits we have to write
var 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 ( var i = 1; i <= n; i *= 10)
number_of_digits += (n - i + 1);
return number_of_digits;
} // Driver code var n = 13;
document.write(totalDigits(n)); </script> |
Output:
17
Time Complexity : O(Logn)