Factorial of Large numbers using Logarithmic identity

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 <bits/stdc++.h>
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 format
string 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 code
int 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 numbers
import 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 numbers
import math
import time

# Maximum number of digits
# in output
MAX=1000

# Function to find the factorial
# of large number and return
# them in string format
def 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 code
tStart = time.perf_counter()
str = factorial(100)
print("The factorial is: ",end="")
print(str)

# Calculates the time taken
# by the algorithm to execute
tEnd = 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 format
function 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 code
const startTime = Date.now();
let str;
str = factorial(100);
console.log("The factorial is: " + str);
// Calculates the time taken
// by the algorithm to execute
console.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