Find the Factorial of a large number

Try it on GfG Practice

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

Examples: 

Input: 100
Output: 933262154439441526816992388562667004-
         907159682643816214685929638952175999-
         932299156089414639761565182862536979-
         208272237582511852109168640000000000-
         00000000000000

Input: 50
Output: 3041409320171337804361260816606476884-
         4377641568960512000000000000

We have discussed a simple program for factorial.

Why conventional way of computing factorial fails for large numbers?

A factorial of 100 has 158 digits. It is not possible to store these many digits even if we use long int.

The idea is to use basic mathematics for multiplication.

Illustration:

Example to show working of multiply(res[], x)

• A number 5189 is stored in res[] as following: res[] = {9, 8, 1, 5}
• let x = 10
  Initialize carry = 0

• At i = 0, prod = res[0]*x + carry = 9*10 + 0 = 90.
  res[0] = 0, carry = 9

 

• At i = 1, prod = res[1]*x + carry = 8*10 + 9 = 89
  res[1] = 9, carry = 8

 

• At i = 2, prod = res[2]*x + carry = 1*10 + 8 = 18
  res[2] = 8, carry = 1

 

• At i = 3, prod = res[3]*x + carry = 5*10 + 1 = 51
  res[3] = 1, carry = 5

 

• res[4] = carry = 5
  res[] = {0, 9, 8, 1, 5} 

Follow the steps below to solve the given problem:

• Create an array res[] of MAX size where MAX is a number of maximum digits in output. 
• Initialize value stored in res[] as 1 and initialize res_size (size of 'res[]') as 1. 
• Multiply x with res[] and update res[] and res_size to store the multiplication result for all the numbers from x = 2 to n.
• To multiply a number x with the number stored in res[], one by one multiply x with every digit of res[].
• To implement multiply function perform the following steps:
  • Initialize carry as 0. 
  • Do following for i = 0 to res_size - 1 
    • Find value of res[i] * x + carry. Let this value be prod. 
    • Update res[i] by storing the last digit of prod in it. 
    • Update carry by storing the remaining digits in carry. 
  • Put all digits of carry in res[] and increase res_size by the number of digits in carry.

Below is the implementation of the above algorithm. 

NOTE: In the below implementation, the maximum digits in the output are assumed as 500. To find a factorial of a much larger number ( > 254), increase the size of an array or increase the value of MAX. This can also be solved using Linked List instead of using res[] array which will not waste extra space.

// C++ program to compute factorial of big numbers
#include <iostream>
using namespace std;

// Maximum number of digits in output
#define MAX 500

int multiply(int x, int res[], int res_size);

// This function finds factorial of large numbers
// and prints them
void factorial(int n)
{
    int res[MAX];

    // Initialize result
    res[0] = 1;
    int res_size = 1;

    // Apply simple factorial formula n! = 1 * 2 * 3
    // * 4...*n
    for (int x = 2; x <= n; x++)
        res_size = multiply(x, res, res_size);

    cout << "Factorial of given number is \n";
    for (int i = res_size - 1; i >= 0; i--)
        cout << res[i];
}

// This function multiplies x with the number
// represented by res[].
// res_size is size of res[] or number of digits in the
// number represented by res[]. This function uses simple
// school mathematics for multiplication.
// This function returns the
// new value of res_size
int multiply(int x, int res[], int res_size)
{
    int carry = 0; // Initialize carry

    // One by one multiply n with individual digits of res[]
    for (int i = 0; i < res_size; i++) {
        int prod = res[i] * x + carry;

        // Store last digit of 'prod' in res[]
        res[i] = prod % 10;

        // Put rest in carry
        carry = prod / 10;
    }

    // Put carry in res and increase result size
    while (carry) {
        res[res_size] = carry % 10;
        carry = carry / 10;
        res_size++;
    }
    return res_size;
}

// Driver program
int main()
{
    factorial(100);
    return 0;
}

// JAVA program to compute factorial
// of big numbers
class GFG {

    // This function finds factorial of
    // large numbers and prints them
    static void factorial(int n)
    {
        int res[] = new int[500];

        // Initialize result
        res[0] = 1;
        int res_size = 1;

        // Apply simple factorial formula
        // n! = 1 * 2 * 3 * 4...*n
        for (int x = 2; x <= n; x++)
            res_size = multiply(x, res, res_size);

        System.out.println("Factorial of given number is ");
        for (int i = res_size - 1; i >= 0; i--)
            System.out.print(res[i]);
    }

    // This function multiplies x with the number
    // represented by res[]. res_size is size of res[] or
    // number of digits in the number represented by res[].
    // This function uses simple school mathematics for
    // multiplication. This function may value of res_size
    // and returns the new value of res_size
    static int multiply(int x, int res[], int res_size)
    {
        int carry = 0; // Initialize carry

        // One by one multiply n with individual
        // digits of res[]
        for (int i = 0; i < res_size; i++) {
            int prod = res[i] * x + carry;
            res[i] = prod % 10; // Store last digit of
                                // 'prod' in res[]
            carry = prod / 10; // Put rest in carry
        }

        // Put carry in res and increase result size
        while (carry != 0) {
            res[res_size] = carry % 10;
            carry = carry / 10;
            res_size++;
        }
        return res_size;
    }

    // Driver program
    public static void main(String args[])
    {
        factorial(100);
    }
}
// This code is contributed by Nikita Tiwari

# Python program to compute factorial
# of big numbers

import sys

# This function finds factorial of large
# numbers and prints them


def factorial(n):
    res = [None]*500
    # Initialize result
    res[0] = 1
    res_size = 1

    # Apply simple factorial formula
    # n! = 1 * 2 * 3 * 4...*n
    x = 2
    while x <= n:
        res_size = multiply(x, res, res_size)
        x = x + 1

    print("Factorial of given number is")
    i = res_size-1
    while i >= 0:
        sys.stdout.write(str(res[i]))
        sys.stdout.flush()
        i = i - 1


# This function multiplies x with the number
# represented by res[]. res_size is size of res[]
# or number of digits in the number represented
# by res[]. This function uses simple school
# mathematics for multiplication. This function
# may value of res_size and returns the new value
# of res_size
def multiply(x, res, res_size):

    carry = 0  # Initialize carry

    # One by one multiply n with individual
    # digits of res[]
    i = 0
    while i < res_size:
        prod = res[i] * x + carry
        res[i] = prod % 10  # Store last digit of
        # 'prod' in res[]
        # make sure floor division is used
        carry = prod//10  # Put rest in carry
        i = i + 1

    # Put carry in res and increase result size
    while (carry):
        res[res_size] = carry % 10
        # make sure floor division is used
        # to avoid floating value
        carry = carry // 10
        res_size = res_size + 1

    return res_size


# Driver program
factorial(100)

# This code is contributed by Nikita Tiwari.

// C# program to compute
// factorial of big numbers
using System;

class GFG {

    // This function finds factorial
    // of large numbers and prints them
    static void factorial(int n)
    {
        int[] res = new int[500];

        // Initialize result
        res[0] = 1;
        int res_size = 1;

        // Apply simple factorial formula
        // n! = 1 * 2 * 3 * 4...*n
        for (int x = 2; x <= n; x++)
            res_size = multiply(x, res, res_size);

        Console.WriteLine("Factorial of "
                          + "given number is ");
        for (int i = res_size - 1; i >= 0; i--)
            Console.Write(res[i]);
    }

    // This function multiplies x
    // with the number represented
    // by res[]. res_size is size
    // of res[] or number of digits
    // in the number represented by
    // res[]. This function uses
    // simple school mathematics for
    // multiplication. This function
    // may value of res_size and
    // returns the new value of res_size
    static int multiply(int x, int[] res, int res_size)
    {
        int carry = 0; // Initialize carry

        // One by one multiply n with
        // individual digits of res[]
        for (int i = 0; i < res_size; i++) {
            int prod = res[i] * x + carry;
            res[i] = prod % 10; // Store last digit of
                                // 'prod' in res[]
            carry = prod / 10; // Put rest in carry
        }

        // Put carry in res and
        // increase result size
        while (carry != 0) {
            res[res_size] = carry % 10;
            carry = carry / 10;
            res_size++;
        }
        return res_size;
    }

    // Driver Code
    static public void Main() { factorial(100); }
}

// This code is contributed by ajit

<script>

// Javascript program to compute factorial of big numbers

// This function finds factorial of large numbers
// and prints them
function factorial(n)
{
    let res = new Array(500);

    // Initialize result
    res[0] = 1;
    let res_size = 1;

    // Apply simple factorial formula n! = 1 * 2 * 3 * 4...*n
    for (let x=2; x<=n; x++)
        res_size = multiply(x, res, res_size);

    document.write("Factorial of given number is " + "<br>");
    for (let i=res_size-1; i>=0; i--)
        document.write(res[i]);
}

// This function multiplies x with the number 
// represented by res[].
// res_size is size of res[] or number of digits in the 
// number represented by res[]. This function uses simple 
// school mathematics for multiplication.
// This function may value of res_size and returns the 
// new value of res_size
function multiply(x, res, res_size)
{
    let carry = 0; // Initialize carry

    // One by one multiply n with individual digits of res[]
    for (let i=0; i<res_size; i++)
    {
        let prod = res[i] * x + carry;

        // Store last digit of 'prod' in res[] 
        res[i] = prod % 10; 

        // Put rest in carry
        carry = Math.floor(prod/10); 
    }

    // Put carry in res and increase result size
    while (carry)
    {
        res[res_size] = carry%10;
        carry = Math.floor(carry/10);
        res_size++;
    }
    return res_size;
}

// Driver program
    factorial(100);

// This  code is contributed by Mayank Tyagi

</script>

<?php
// PHP program to compute factorial 
// of big numbers

// Maximum number of digits in output
$MAX = 500;

// This function finds factorial of 
// large numbers and prints them
function factorial($n)
{
    global $MAX;
    $res = array_fill(0, $MAX, 0);

    // Initialize result
    $res[0] = 1;
    $res_size = 1;

    // Apply simple factorial formula 
    // n! = 1 * 2 * 3 * 4...*n
    for ($x = 2; $x <= $n; $x++)
        $res_size = multiply($x, $res, $res_size);

    echo "Factorial of given number is \n";
    for ($i = $res_size - 1; $i >= 0; $i--)
        echo $res[$i];
}

// This function multiplies x with the number 
// represented by res[].
// res_size is size of res[] or number of 
// digits in the number represented by res[]. 
// This function uses simple school mathematics 
// for multiplication. This function may value  
// of res_size and returns the new value of res_size
function multiply($x, &$res, $res_size)
{
    $carry = 0; // Initialize carry

    // One by one multiply n with individual
    // digits of res[]
    for ($i = 0; $i < $res_size; $i++)
    {
        $prod = $res[$i] * $x + $carry;

        // Store last digit of 'prod' in res[] 
        $res[$i] = $prod % 10; 

        // Put rest in carry
        $carry = (int)($prod / 10); 
    }

    // Put carry in res and increase 
    // result size
    while ($carry)
    {
        $res[$res_size] = $carry % 10;
        $carry = (int)($carry / 10);
        $res_size++;
    }
    return $res_size;
}

// Driver Code
factorial(100);
    
// This code is contributed by chandan_jnu
?>

Factorial of given number is 
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

Time Complexity: O(N log (N!)), where O(N) is for loop and O(log N!) is for nested while loop
Auxiliary Space: O(max(digits in factorial))

Find the Factorial of a large number using Basic BigInteger

This problem can be solved using the below idea:

Big Integer can also be used to calculate the factorial of large numbers.

Illustration:

N = 5

ans = 1

At i = 2: ans = ans x i = 1 x 2 = 2
At i = 3: ans = ans x i = 2 x 3 = 6
At i = 4: ans = ans x i = 6 x 4 = 24
At i = 5: ans = ans x i = 24 x 5 = 120

Hence factorial of N is 120

Follow the steps below to solve the given problem: 

• Declare a BigInteger f with 1 and perform the conventional way of calculating factorial
• Traverse a loop from x = 2 to N and multiply x with f and store the resultant value in f

Below is the implementation of the above idea : 

// C++ program to find large
// factorials using BigInteger
#include <bits/stdc++.h>
using namespace std;

#define ull unsigned long long

// Returns Factorial of N
ull factorial(int N)
{

    // Initialize result
    ull f = 1; // Or BigInt 1

    // Multiply f with 2, 3, ...N
    for (ull i = 2; i <= N; i++)
        f *= i;

    return f;
}

// Driver method
int main()
{
    int N = 20;
    cout << factorial(N) << endl;
}

// This code is contributed by phasing17

// Java program to find large
// factorials using BigInteger
import java.math.BigInteger;
import java.util.Scanner;

public class Example {

    // Returns Factorial of N
    static BigInteger factorial(int N)
    {
        // Initialize result
        BigInteger f
            = new BigInteger("1"); // Or BigInteger.ONE

        // Multiply f with 2, 3, ...N
        for (int i = 2; i <= N; i++)
            f = f.multiply(BigInteger.valueOf(i));

        return f;
    }

    // Driver method
    public static void main(String args[]) throws Exception
    {
        int N = 20;
        System.out.println(factorial(N));
    }
}

# Python3 program to find large
# factorials

# Returns Factorial of N
def factorial(N):

    # Initialize result
    f =  1  
    
    # Multiply f with 2, 3, ...N
    for i in range(2, N + 1):
        f *= i
    
    return f;


# Driver method
N = 20;
print(factorial(N));


# This code is contributed by phasing17
    

// C# program to find large
// factorials using BigInteger
using System;
using System.Collections.Generic;
using System.Numerics;

public class Example {

  // Returns Factorial of N
  static BigInteger factorial(int N)
  {

    // Initialize result
    BigInteger f
      = new BigInteger(1); // Or BigInteger.ONE

    // Multiply f with 2, 3, ...N
    for (int i = 2; i <= N; i++)
      f = BigInteger.Multiply(f, new BigInteger(i));

    return f;
  }

  // Driver method
  public static void Main(string[] args)
  {
    int N = 20;
    Console.WriteLine(factorial(N));
  }
}

// This code is contributed by phasing17

// JavaScript program to find large
// factorials using BigInteger

// Returns Factorial of N
function factorial(N)
{

// Initialize result
let f =  BigInt(1); // Or BigInt 1

// Multiply f with 2, 3, ...N
for (var i = 2; i <= N; i++)
    f *= BigInt(i);

return f;
}

// Driver method
let N = 20;
console.log(factorial(N));

// This code is contributed by phasing17
    

2432902008176640000

Time Complexity: O(N)
Auxiliary Space: O(1) 

Find Factorial of a number using Linked List :-

So the basic idea is to multiply the next number ranging between 2 to N with the data stored in the current Node and also maintain a carry just like the array approach. And then move on to the next node.

Let's break it down into following steps :

1. Initially we'll create 1 single node containing 1 in it.
2. Then initialized i of a for loop with 2.
3. And for each value of i up till N, we'll call a function multiply which takes 2 parameter, head of the list and value of i.
4. And perform the below operation (See the image)

The above operation will be carried out till our temp pointer becomes NULL.

Now there are further 2 cases , The multiplication of i with the node's data :

• Doesn't exceed 1 digit
• Exceeds 1 digit

If the multiplication of current node's data with i - Exceeds 1 digit, then we won't be storing it into a single node. Rather, we'll be storing each digit into a single node. And if it doesn't then we'll simply replace current node's data with it.

So, by above explanation we can conclude and form the following steps to solve this problem :

Algorithm:

1. Find value of : node's data * i + carry. Store it in a variable 'prod'.
2. Initialize 2 pointers let's say prev and temp on the current node and Update the current node's value by storing the last digit of the 'prod'
3. Update carry by storing the remaining digits in carry (excluding the last digit)
4. Now, bring prev ptr on temp and move temp to the next node (if any) and then again perform, the previous steps until the carry becomes 0 or the temp ptr becomes NULL.
5. Once the temp pointer reaches the NULL there is a possibility that carry is still not 0. So, until carry becomes 0 we have to again perform the same steps by creating a new node for every remaining digit.

See the image below for the Dry run of above steps :

(i) Node's data = 1, i = 2, carry = 0.

(ii) Node's Data : 2, i = 3, carry = 0

(iii) Node's Data : 6, i = 4, carry = 2 (6*4 = 24, 24 % 10 = 4, 24/10 = 2)

X = NULL in the below image :

Just like this, we'll do for the remaining digits, and each of the digit will be stored in a single node.

Code :

#include <bits/stdc++.h>
using namespace std;

//* Node Class
class Node {
public:
    int data;
    Node* prev;
    Node(int n)
    {
        data = n;
        prev = NULL;
    }
};

//* Function to perform desired operation
void Multiply(Node* head, int i)
{
    Node *temp = head,
        *prevPtr = head; // Temp variable for keeping head
    
    int carry = 0;

    //* Perform operation until temp becomes NULL
    while (temp != NULL) {
        int prod = temp->data * i + carry;
        temp->data = prod % 10; //* Stores the last digit
        carry = prod / 10;
        prevPtr = temp; //* Change Links
        temp = temp->prev; //* Moving temp to next node
    }

    //* If carry is greater than 0 then we create new nodes
    //* to store remaining digits.
    while (carry != 0) {
        prevPtr->prev = new Node((int)(carry % 10));
        carry /= 10;
        prevPtr = prevPtr->prev;
    }
}

//* Using head recursion to print the linked list's data in reverse
void print(Node* head)
{
    if (head == NULL) 
        return;
    print(head->prev);
    cout << head->data; // Print linked list in reverse order
}

// Driver code
int main()
{
    int n = 100;
    Node *head = new Node(1); // Create a node and initialise it by 1
    
    for(int i = 2; i <= n; i++)
        Multiply(head, i); // Run a loop from 2 to n and
                            // multiply with head's i
    cout << "Factorial of " << n << " is : \n";
    print(head); // Print the linked list
    cout << endl;
    return 0;
}

//* Node Class
class Node {
    public int data;
    public Node prev;

    public Node(int n) {
        data = n;
        prev = null;
    }
}

//* Function to perform desired operation
class Main {
    public static void Multiply(Node head, int i) {
        Node temp = head;
        Node prevPtr = head; // Temp variable for keeping head
        int carry = 0;

        //* Perform operation until temp becomes null
        while (temp != null) {
            int prod = temp.data * i + carry;
            temp.data = prod % 10; //* Stores the last digit
            carry = prod / 10;
            prevPtr = temp; //* Change Links
            temp = temp.prev; //* Moving temp to next node
        }

        //* If carry is greater than 0 then we create new nodes
        //* to store remaining digits.
        while (carry != 0) {
            prevPtr.prev = new Node((int) (carry % 10));
            carry /= 10;
            prevPtr = prevPtr.prev;
        }
    }

    //* Using head recursion to print the linked list's data in reverse
    public static void print(Node head) {
        if (head == null)
            return;
        print(head.prev);
        System.out.print(head.data); // Print linked list in reverse order
    }

    // Driver code
    public static void main(String[] args) {
        int n = 100;
        Node head = new Node(1); // Create a node and initialize it by 1

        for (int i = 2; i <= n; i++)
            Multiply(head, i); // Run a loop from 2 to n and multiply with head's i

        System.out.println("Factorial of " + n + " is : ");
        print(head); // Print the linked list
        System.out.println();
    }
}

// by phasing17

# Node Class
class Node:
    def __init__(self, n):
        self.data = n
        self.prev = None

# Function to perform desired operation
def Multiply(head, i):
    temp = head
    prevPtr = head
    carry = 0
    # Perform operation until temp becomes None
    while temp is not None:
        prod = temp.data * i + carry
        temp.data = prod % 10 # Stores the last digit
        carry = prod // 10
        prevPtr = temp # Change Links
        temp = temp.prev # Moving temp to the next node
    # If carry is greater than 0, create new nodes to store remaining digits
    while carry != 0:
        prevPtr.prev = Node(carry % 10)
        carry = carry // 10
        prevPtr = prevPtr.prev

# Using recursion to print the linked list's data in reverse
def print_list(head):
    if head is None:
        return
    print_list(head.prev)
    print(head.data, end="") # Print linked list in reverse order

# Driver code
def main():
    n = 100
    head = Node(1) # Create a node and initialize it by 1
    for i in range(2, n+1):
        Multiply(head, i) # Run a loop from 2 to n and multiply with head's i
    print("Factorial of", n, "is : ")
    print_list(head) # Print the linked list
    print()

main()

using System;

// Node Class
public class Node
{
    public int data;
    public Node prev;
    
    public Node(int n)
    {
        data = n;
        prev = null;
    }
}

// Function to perform desired operation
public class Program
{
    public static void Multiply(Node head, int i)
    {
        Node temp = head;
        Node prevPtr = head;
        int carry = 0;

        // Perform operation until temp becomes null
        while (temp != null)
        {
            int prod = temp.data * i + carry;
            temp.data = prod % 10;
            carry = prod / 10;
            prevPtr = temp;
            temp = temp.prev;
        }

        // If carry is greater than 0 then we create new nodes
        // to store remaining digits.
        while (carry != 0)
        {
            prevPtr.prev = new Node((int)(carry % 10));
            carry /= 10;
            prevPtr = prevPtr.prev;
        }
    }

    // Using head recursion to print the linked list's data in reverse
    public static void Print(Node head)
    {
        if (head == null)
            return;
        Print(head.prev);
        Console.Write(head.data); // Print linked list in reverse order
    }

    // Driver code
    public static void Main()
    {
        int n = 100;
        Node head = new Node(1); // Create a node and initialize it by 1

        for (int i = 2; i <= n; i++)
            Multiply(head, i); // Run a loop from 2 to n and
                              // multiply with head's i

        Console.WriteLine("Factorial of " + n + " is : ");
        Print(head); // Print the linked list
        Console.WriteLine();
    }
}

// Node Class
class Node {
    constructor(n) {
        this.data = n;
        this.prev = null;
    }
}

// Function to perform desired operation
function Multiply(head, i) {
    let temp = head;
    let prevPtr = head;
    let carry = 0;

    // Perform operation until temp becomes null
    while (temp !== null) {
        let prod = temp.data * i + carry;
        temp.data = prod % 10; // Stores the last digit
        carry = Math.floor(prod / 10);
        prevPtr = temp; // Change Links
        temp = temp.prev; // Moving temp to the next node
    }

    // If carry is greater than 0, create new nodes to store remaining digits
    while (carry !== 0) {
        prevPtr.prev = new Node(carry % 10);
        carry = Math.floor(carry / 10);
        prevPtr = prevPtr.prev;
    }
}

// Using recursion to print the linked list's data in reverse
function print(head) {
    if (head === null)
        return;
    print(head.prev);
    process.stdout.write(head.data.toString()); // Print linked list in reverse order
}

// Driver code
function main() {
    const n = 100;
    const head = new Node(1); // Create a node and initialize it by 1

    for (let i = 2; i <= n; i++)
        Multiply(head, i); // Run a loop from 2 to n and multiply with head's i

    console.log("Factorial of " + n + " is : ");
    print(head); // Print the linked list
    console.log();
}

main();

Factorial of 100 is : 
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

Time Complexity : O(N²)
Space Complexity : O(digits in factorial)