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# Count digits in a factorial using Logarithm

• Difficulty Level : Medium
• Last Updated : 31 Mar, 2023

Given an integer N, 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:  5
Output: 3
Explanation: 5! = 120, i.e., 3 digits

Input: 10
Output: 7
Explanation: 10! = 3628800, i.e., 7 digits

Naive approach: To solve the problem follow the below idea:

A naive solution would be to calculate the n! first and then calculate the number of digits present in it. However as the value for n! can be very large, it would become cumbersome to store them in a variable (Unless you’re working in python!).

## Count digits in a factorial using the property of logarithms:

To solve the problem follow the below idea:

We know,
log(a*b) = log(a) + log(b)

Therefore
log( n! ) = log(1*2*3……. * n) = log(1) + log(2) + …….. +log(n)

Now, observe that the floor value of log base
10 increased by 1, of any number, gives the
number of digits present in that number.
Hence, output would be : floor(log(n!)) + 1.

Below is the implementation of the above approach:

## C++

 `// A C++ program to find the number of digits in``// a factorial` `#include ``using` `namespace` `std;` `// This function receives an integer n, and returns``// the number of digits present in n!``int` `findDigits(``int` `n)``{``    ``// factorial exists only for n>=0``    ``if` `(n < 0)``        ``return` `0;` `    ``// base case``    ``if` `(n <= 1)``        ``return` `1;` `    ``// else iterate through n and calculate the``    ``// value``    ``double` `digits = 0;``    ``for` `(``int` `i = 2; i <= n; i++)``        ``digits += ``log10``(i);` `    ``return` `floor``(digits) + 1;``}` `// Driver code``int` `main()``{``      ``// Function call``    ``cout << findDigits(1) << endl;``    ``cout << findDigits(5) << endl;``    ``cout << findDigits(10) << endl;``    ``cout << findDigits(120) << endl;``    ``return` `0;``}`

## Java

 `// Java program to find the number``// of digits in a factorial` `import` `java.io.*;``import` `java.util.*;` `class` `GFG {``    ``// returns the number of digits``    ``// present in n!``    ``static` `int` `findDigits(``int` `n)``    ``{``        ``// factorial exists only for n>=0``        ``if` `(n < ``0``)``            ``return` `0``;` `        ``// base case``        ``if` `(n <= ``1``)``            ``return` `1``;` `        ``// else iterate through n and calculate the``        ``// value``        ``double` `digits = ``0``;``        ``for` `(``int` `i = ``2``; i <= n; i++)``            ``digits += Math.log10(i);` `        ``return` `(``int``)(Math.floor(digits)) + ``1``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``          ``// Function call``        ``System.out.println(findDigits(``1``));``        ``System.out.println(findDigits(``5``));``        ``System.out.println(findDigits(``10``));``        ``System.out.println(findDigits(``120``));``    ``}``}` `// This code is contributed by Pramod Kumar`

## Python3

 `# Python3 program to find the``# number of digits in a factorial``import` `math` `# This function receives an integer``# n, and returns the number of``# digits present in n!`  `def` `findDigits(n):` `    ``# factorial exists only for n>=0``    ``if` `(n < ``0``):``        ``return` `0` `    ``# base case``    ``if` `(n <``=` `1``):``        ``return` `1` `    ``# else iterate through n and``    ``# calculate the value``    ``digits ``=` `0``    ``for` `i ``in` `range``(``2``, n ``+` `1``):``        ``digits ``+``=` `math.log10(i)` `    ``return` `math.floor(digits) ``+` `1`  `# Driver code``if` `__name__ ``=``=` `"__main__"``:``  ``print``(findDigits(``1``))``  ``print``(findDigits(``5``))``  ``print``(findDigits(``10``))``  ``print``(findDigits(``120``))` `# This code is contributed by mits`

## Javascript

 `// A Javascript program to find the number of digits in``// a factorial`  `// This function receives an integer n, and returns``// the number of digits present in n!``function` `findDigits(n)``{``    ``// factorial exists only for n>=0``    ``if` `(n < 0)``        ``return` `0;` `    ``// base case``    ``if` `(n <= 1)``        ``return` `1;` `    ``// else iterate through n and calculate the``    ``// value``    ``let digits = 0;``    ``for` `(let i=2; i<=n; i++)``        ``digits += Math.log10(i);` `    ``return` `Math.floor(digits) + 1;``}` `// Driver code`` ` `    ``document.write(findDigits(1) + ``"
"``);``    ``document.write(findDigits(5) + ``"
"``);``    ``document.write(findDigits(10) + ``"
"``);``    ``document.write(findDigits(120) + ``"
"``);``    `  `//This code is contributed by Mayank Tyagi`

## C#

 `// A C# program to find the number``// of digits in a factorial``using` `System;` `class` `GFG {` `    ``// This function receives an integer``    ``// n, and returns the number of``    ``// digits present in n!``    ``static` `int` `findDigits(``int` `n)``    ``{` `        ``// factorial exists only for n>=0``        ``if` `(n < 0)``            ``return` `0;` `        ``// base case``        ``if` `(n <= 1)``            ``return` `1;` `        ``// else iterate through n and``        ``// calculate the value``        ``double` `digits = 0;``        ``for` `(``int` `i = 2; i <= n; i++)``            ``digits += Math.Log10(i);` `        ``return` `(``int``)Math.Floor(digits) + 1;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``          ``// Function call``        ``Console.Write(findDigits(1) + ``"\n"``);``        ``Console.Write(findDigits(5) + ``"\n"``);``        ``Console.Write(findDigits(10) + ``"\n"``);``        ``Console.Write(findDigits(120) + ``"\n"``);``    ``}``}` `// This code is contributed by``// Smitha Dinesh Semwal`

## PHP

 `=0``    ``if` `(``\$n` `< 0)``        ``return` `0;` `    ``// base case``    ``if` `(``\$n` `<= 1)``        ``return` `1;` `    ``// else iterate through n and``    ``// calculate the value``    ``\$digits` `= 0;``    ``for` `(``\$i` `= 2; ``\$i` `<= ``\$n``; ``\$i``++)``        ``\$digits` `+= log10(``\$i``);` `    ``return` `floor``(``\$digits``) + 1;``}` `// Driver code` `// Function call``echo` `findDigits(1), ``"\n"``;``echo` `findDigits(5), ``"\n"``;``echo` `findDigits(10), ``"\n"``;``echo` `findDigits(120), ``"\n"``;` `// This code is contributed by Ajit.``?>`

Output

```1
3
7
199```

Auxiliary space: O(1) because it is using constant variables

Approach 2: Using Stirling’s approximation formula to calculate the factorial and logarithm to count the number of digits.

1. The countDigitsInFactorial(int n) function takes an integer n as input and returns the number of digits in the factorial of n. If n is negative, it returns 0. If n is 0 or 1, the factorial is 1, and it returns 1.
2. In the countDigitsInFactorial(int n) function, the double x variable is declared and initialized using the Stirling’s approximation formula for the factorial. This formula provides a good approximation of the value of the factorial for large values of n.
3. where e is the mathematical constant, and π is the mathematical constant pi.
4. The formula used in this code is a simplified version of Stirling’s approximation that takes the logarithm of the above formula to get the number of digits in the factorial.

## C++

 `#include ``#include ``using` `namespace` `std;` `int` `countDigitsInFactorial(``int` `n)``{``    ``if` `(n < 0) {``        ``return` `0;``    ``}``    ``if` `(n <= 1) {``        ``return` `1;``    ``}``    ``double` `x``        ``= (n * ``log10``(n / M_E) + ``log10``(2 * M_PI * n) / 2.0);``    ``return` `floor``(x) + 1;``}` `int` `main()``{``    ``cout << countDigitsInFactorial(1) << endl;``    ``cout << countDigitsInFactorial(5) << endl;``    ``cout << countDigitsInFactorial(10) << endl;``    ``cout << countDigitsInFactorial(120) << endl;` `    ``return` `0;``}`

## Java

 `// Java implementation of above approach``import` `java.lang.Math;``import` `java.util.Scanner;` `public` `class` `Solution {``public` `static` `int` `countDigitsInFactorial(``int` `n) {``  ` ` ``// factorial exists only for n>=0``if` `(n < ``0``) {``return` `0``;``}``if` `(n <= ``1``) {``return` `1``;``}``  ``// Calculating the digit's values``double` `x = (n * Math.log10(n / Math.E) + Math.log10(``2` `* Math.PI * n) / ``2.0``);``  ``// returning the floor value + 1``return` `(``int``) Math.floor(x) + ``1``;``}` `public` `static` `void` `main(String[] args) {``  ` `    ``// calling the countDigitInFactorial function``    ``System.out.println(countDigitsInFactorial(``1``));``    ``System.out.println(countDigitsInFactorial(``5``));``    ``System.out.println(countDigitsInFactorial(``10``));``    ``System.out.println(countDigitsInFactorial(``120``));``}``}`

## Python3

 `import` `math`  `def` `countDigitsInFactorial(n):``    ``if` `n < ``0``:``        ``return` `0``    ``if` `n <``=` `1``:``        ``return` `1` `    ``# Using Stirling's approximation formula to count the number of digits``    ``x ``=` `(n ``*` `math.log10(n ``/` `math.e) ``+` `math.log10(``2` `*` `math.pi ``*` `n) ``/` `2.0``)` `    ``# Floor the result of the formula and add 1 to get the number of digits``    ``return` `math.floor(x) ``+` `1`  `# Testing the function with sample inputs``print``(countDigitsInFactorial(``1``))``print``(countDigitsInFactorial(``5``))``print``(countDigitsInFactorial(``10``))``print``(countDigitsInFactorial(``120``))`

Output

```1
3
7
199```

Time complexity: O(1)

The time complexity of the above approach to count the number of digits in n! using Stirling’s approximation and logarithms is O(1), meaning it is constant time complexity.
Auxiliary space: O(1)

In the next set, we’d see how to further optimize our approach and reduce the time complexity for the same program.
This article is contributed by Ashutosh Kumar . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.