# Factorial of Large numbers using Logarithmic identity

Last Updated : 07 Jan, 2024

Given a very large number N, the task is to find the factorial of the number using Log.

Factorial of a non-negative integer is the multiplication of all integers smaller than or equal to N.

We have previously discussed a simple program to find the factorial in this article. Here, we will discuss an efficient way to find the factorial of large numbers. Examples:

Input: N = 100
Output: 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

Input: N = 50
Output: 30414093201713378043612608166064768844377641568960512000000000000

Approach: The most common iterative version runs in expected O(N) time. But as numbers become big it will be wrong to assume that multiplication takes constant time. The naive approach takes O(K*M) time for multiplication where K is the length of the multiplier and M is the length of the multiplicand. Therefore, the idea is to use logarithmic properties: As we know that and Therefore: Another property is by substituting the value of ln(N!). Below is the implementation of the above approach:

## C++

 // C++ program to compute the// factorial of big numbers #include using namespace std; // Maximum number of digits// in output#define MAX 1000 // Function to find the factorial// of large number and return// them in string formatstring factorial(long long n){    if (n > MAX) {        cout << " Integer Overflow"             << endl;        return "";    }     long long counter;    long double sum = 0;     // Base case    if (n == 0)        return "1";     // Calculate the sum of    // logarithmic values     for (counter = 1; counter <= n;         counter++) {        sum = sum + log(counter);    }     // Number becomes too big to hold in    // unsigned long integers.    // Hence converted to string    // Answer is sometimes under    // estimated due to floating point    // operations so round() is used    string result        = to_string(round(exp(sum)));     return result;} // Driver codeint main(){    clock_t tStart = clock();    string str;    str = factorial(100);    cout << "The factorial is: "         << str << endl;     // Calculates the time taken    // by the algorithm to execute    cout << "Time taken: " << setprecision(10)         << ((double)(clock() - tStart)             / CLOCKS_PER_SEC)         << " s" << endl;}

## Java

 // Java program to compute the// factorial of big numbersimport java.math.BigDecimal;import java.math.RoundingMode; public class Factorial {// Maximum number of digits// in output  static int MAX = 1000; // Function to find the factorial// of large number and return// them in string format  static String factorial(int n) {    if (n > MAX) {      System.out.println(" Integer Overflow");      return "";    }     int counter;    double sum = 0;     // Base case    if (n == 0)      return "1";     // Calculate the sum of    // logarithmic values    for (counter = 1; counter <= n; counter++) {      sum = sum + Math.log(counter);    }     // Number becomes too big to hold in    // unsigned long integers.    // Hence converted to string    // Answer is sometimes under    // estimated due to floating point    // operations so round() is used    BigDecimal result = new BigDecimal(Math.exp(sum));    result = result.setScale(0, RoundingMode.HALF_UP);     return result.toString();  }     // Driver code  public static void main(String[] args) {    long startTime = System.currentTimeMillis();    String str;    str = factorial(100);    System.out.println("The factorial is: " + str);          // Calculates the time taken    // by the algorithm to execute    System.out.println("Time taken: " + (System.currentTimeMillis() - startTime) + " ms");  }} // This code is contributed by Aman Kumar.

## Python3

 # Python program to compute the# factorial of big numbersimport mathimport time # Maximum number of digits# in outputMAX=1000 # Function to find the factorial# of large number and return# them in string formatdef factorial(n):    if(n > MAX):        print(" Integer Overflow")        return ""         counter = 0    sum = 0         # Base case    if(n == 0):        return "1"             # Calculate the sum of     # logarithmic values         for counter in range(1,n+1):        sum = sum + math.log(counter)             # Number becomes too big to hold in    # unsigned long integers.    # Hence converted to string    # Answer is sometimes under    # estimated due to floating point    # operations so round() is used    result = str(round(math.exp(sum)))         return result     # Driver codetStart = time.perf_counter()str = factorial(100)print("The factorial is: ",end="")print(str) # Calculates the time taken# by the algorithm to executetEnd = time.perf_counter()print("Time taken: ",end="")print('%.10f' % (tEnd-tStart),end="")print(" s") # This code is contributed by Pushpesh Raj.

## Javascript

 // Javascript program to compute the// factorial of big numbers // Function to find the factorial// of large number and return// them in string formatfunction factorial(n) {    const MAX = 1000;    if (n > MAX) {        console.log(" Integer Overflow");        return "";    }    let counter;    let sum = 0;    // Base case    if (n === 0)        return "1";    // Calculate the sum of    // logarithmic values    for (counter = 1; counter <= n; counter++) {        sum = sum + Math.log(counter);    }    // Number becomes too big to hold in    // unsigned long integers.    // Hence converted to string    // Answer is sometimes under    // estimated due to floating point    // operations so round() is used    const result = BigInt(Math.round(Math.exp(sum))).toString();    return result;} // Driver codeconst startTime = Date.now();let str;str = factorial(100);console.log("The factorial is: " + str);// Calculates the time taken// by the algorithm to executeconsole.log("Time taken: " + (Date.now() - startTime) + " ms");

## C#

 using System;using System.Diagnostics;using System.Numerics; public class Factorial {    const int MAX = 1000;     static string ComputeFactorial(long n) {        if (n > MAX) {            Console.WriteLine(" Integer Overflow");            return "";        }         long counter;        double sum = 0;         if (n == 0)            return "1";         for (counter = 1; counter <= n; counter++) {            sum = sum + Math.Log(counter);        }         string result = BigInteger.Exp(sum).ToString();         return result;    }     static void Main(string[] args) {        Stopwatch stopwatch = Stopwatch.StartNew();        string str;        str = ComputeFactorial(100);        Console.WriteLine("The factorial is: " + str);        Console.WriteLine("Time taken: " + stopwatch.Elapsed.TotalSeconds.ToString("0.000000") + " s");    }}

Output
The factorial is: 93326215443944231979346762015249956831505959550546075483971433508015162170687116519232751238036777284091181469944786448222582618323317549251483571058789842944.000000
Time taken: 0.000198 s

Time Complexity: O(N), where N is the given number.
Auxiliary Space: O(1) since using constant variables

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