Count digits in a factorial | Set 2
Given an integer n (can be very large), find the number of digits that appear in its factorial, where factorial is defined as, factorial(n) = 1*2*3*4……..*n and factorial(0) = 1
Examples:
Input : n = 1 Output : 1 1! = 1, hence number of digits is 1 Input : 5 Output : 3 5! = 120, i.e., 3 digits Input : 10 Output : 7 10! = 3628800, i.e., 7 digits Input : 50000000 Output : 363233781 Input : 1000000000 Output : 8565705523
We’ve already discussed the solution for small values of n in the Count digits in a factorial | Set 1. However that solution would not be able to handle cases where n >10^6
So, can we improve our solution ?
Yes ! we can.
We can use Kamenetsky’s formula to find our answer !
It approximates the number of digits in a factorial by : f(x) = log10( ((n/e)^n) * sqrt(2*pi*n)) Thus, we can pretty easily use the property of logarithms to, f(x) = n* log10(( n/ e)) + log10(2*pi*n)/2
And that’s it !
Our solution can handle very large inputs that can be accommodated in a 32 bit integer,
and even beyond that ! .
Below is the implementation of above idea :
C++
// A optimised program to find the
// number of digits in a factorial
#include <bits/stdc++.h>
using namespace std;
// Returns the number of digits present
// in n! Since the result can be large
// long long is used as return type
long long findDigits(int n)
{
// factorial of -ve number
// doesn't exists
if (n < 0)
return 0;
// base case
if (n <= 1)
return 1;
// Use Kamenetsky formula to calculate
// the number of digits
double x = ((n * log10(n / M_E) +
log10(2 * M_PI * n) /
2.0));
return floor(x) + 1;
}
// Driver Code
int main()
{
cout << findDigits(1) << endl;
cout << findDigits(50000000) << endl;
cout << findDigits(1000000000) << endl;
cout << findDigits(120) << endl;
return 0;
}
Java
// An optimised java program to find the
// number of digits in a factorial
import java.io.*;
import java.util.*;
class GFG {
public static double M_E = 2.71828182845904523536;
public static double M_PI = 3.141592654;
// Function returns the number of
// digits present in n! since the
// result can be large, long long
// is used as return type
static long findDigits(int n)
{
// factorial of -ve number doesn't exists
if (n < 0)
return 0;
// base case
if (n <= 1)
return 1;
// Use Kamenetsky formula to calculate
// the number of digits
double x = (n * Math.log10(n / M_E) +
Math.log10(2 * M_PI * n) /
2.0);
return (long)Math.floor(x) + 1;
}
// Driver Code
public static void main(String[] args)
{
System.out.println(findDigits(1));
System.out.println(findDigits(50000000));
System.out.println(findDigits(1000000000));
System.out.println(findDigits(120));
}
}
// This code is contributed by Pramod Kumar.
Python3
# A optimised Python3 program to find
# the number of digits in a factorial
import math
# Returns the number of digits present
# in n! Since the result can be large
# long long is used as return type
def findDigits(n):
# factorial of -ve number
# doesn't exists
if (n < 0):
return 0;
# base case
if (n <= 1):
return 1;
# Use Kamenetsky formula to
# calculate the number of digits
x = ((n * math.log10(n / math.e) +
math.log10(2 * math.pi * n) /2.0));
return math.floor(x) + 1;
# Driver Code
print(findDigits(1));
print(findDigits(50000000));
print(findDigits(1000000000));
print(findDigits(120));
# This code is contributed by mits
C#
// An optimised C# program to find the
// number of digits in a factorial.
using System;
class GFG {
public static double M_E = 2.71828182845904523536;
public static double M_PI = 3.141592654;
// Function returns the number of
// digits present in n! since the
// result can be large, long long
// is used as return type
static long findDigits(int n)
{
// factorial of -ve number
// doesn't exists
if (n < 0)
return 0;
// base case
if (n <= 1)
return 1;
// Use Kamenetsky formula to calculate
// the number of digits
double x = (n * Math.Log10(n / M_E) +
Math.Log10(2 * M_PI * n) /
2.0);
return (long)Math.Floor(x) + 1;
}
// Driver Code
public static void Main()
{
Console.WriteLine(findDigits(1));
Console.WriteLine(findDigits(50000000));
Console.WriteLine(findDigits(1000000000));
Console.Write(findDigits(120));
}
}
// This code is contributed by Nitin Mittal
PHP
<?php
// A optimised PHP program to find the
// number of digits in a factorial
// Returns the number of digits present
// in n! Since the result can be large
// long long is used as return type
function findDigits($n)
{
// factorial of -ve number
// doesn't exists
if ($n < 0)
return 0;
// base case
if ($n <= 1)
return 1;
// Use Kamenetsky formula to
// calculate the number of digits
$x = (($n * log10($n / M_E) +
log10(2 * M_PI * $n) /
2.0));
return floor($x) + 1;
}
// Driver Code
echo findDigits(1)."\n" ;
echo findDigits(50000000)."\n" ;
echo findDigits(1000000000)."\n" ;
echo findDigits(120) ;
// This code is contributed by nitin mittal
?>
Javascript
<script>
// A optimised Javascript program to find the
// number of digits in a factorial
// Returns the number of digits present
// in n! Since the result can be large
// long long is used as return type
function findDigits(n)
{
// factorial of -ve number
// doesn't exists
if (n < 0)
return 0;
// base case
if (n <= 1)
return 1;
// Use Kamenetsky formula to calculate
// the number of digits
let x = ((n * Math.log10(n / Math.E) +
Math.log10(2 * Math.PI * n) /
2.0));
return Math.floor(x) + 1;
}
// Driver Code
document.write(findDigits(1) + "<br>");
document.write(findDigits(50000000) + "<br>");
document.write(findDigits(1000000000) + "<br>");
document.write(findDigits(120) + "<br>");
// This code is contributed by Mayank Tyagi
</script>
Output:
1 363233781 8565705523 199
References : oeis.org
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