Find the Factorial of a large number (original) (raw)

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

Factorial of a number

**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 onemultiply 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++ `

// C++ program to compute factorial of big numbers #include 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

// 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

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#

// 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

JavaScript

PHP

`

Output

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 1and 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++ `

// 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

// 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));
}

}

Python

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#

// 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

// 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

`

Output

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)

Operation

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 :

6 × 4 = 24

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

Code :

C++ `

#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;

}

Java

//* 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

Python

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()

C#

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();
}

}

JavaScript

// 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();

`

Output

Factorial of 100 is : 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

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