Find the factorial of a large number.
What is Factorial of a number?
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-
00000000000000Input: 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++
// C++ program to compute factorial of big numbers#include <iostream>using namespace std;// Maximum number of digits in output#define MAX 500int multiply(int x, int res[], int res_size);// This function finds factorial of large numbers// and prints themvoid 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_sizeint 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 programint main(){ factorial(100); return 0;} |
Java
// JAVA program to compute factorial// of big numbersclass 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 |
Python3
# Python program to compute factorial# of big numbersimport sys# This function finds factorial of large# numbers and prints themdef 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_sizedef 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 programfactorial(100)# This code is contributed by Nikita Tiwari. |
C#
// C# program to compute// factorial of big numbersusing 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
<script>// Javascript program to compute factorial of big numbers// This function finds factorial of large numbers// and prints themfunction 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_sizefunction 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// PHP program to compute factorial // of big numbers// Maximum number of digits in output$MAX = 500;// This function finds factorial of // large numbers and prints themfunction factorial($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_sizefunction 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 Codefactorial(100); // This code is contributed by chandan_jnu?> |
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 = 120Hence 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++
// C++ program to find large// factorials using BigInteger#include <bits/stdc++.h>using namespace std;#define ull unsigned long long// Returns Factorial of Null 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 methodint main(){ int N = 20; cout << factorial(N) << endl;}// This code is contributed by phasing17 |
Java
// Java program to find large// factorials using BigIntegerimport 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
# Python3 program to find large# factorials# Returns Factorial of Ndef factorial(N): # Initialize result f = 1 # Multiply f with 2, 3, ...N for i in range(2, N + 1): f *= i return f;# Driver methodN = 20;print(factorial(N));# This code is contributed by phasing17 |
C#
// C# program to find large// factorials using BigIntegerusing 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 Nfunction factorial(N){// Initialize resultlet f = BigInt(1); // Or BigInt 1// Multiply f with 2, 3, ...Nfor (var i = 2; i <= N; i++) f *= BigInt(i);return f;}// Driver methodlet 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 :
- Initially we’ll create 1 single node containing 1 in it.
- Then initialized i of a for loop with 2.
- 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.
- 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:
- Find value of : node’s data * i + carry. Store it in a variable ‘prod’.
- 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’
- Update carry by storing the remaining digits in carry (excluding the last digit)
- 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.
- 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 Classclass Node {public: int data; Node* prev; Node(int n) { data = n; prev = NULL; }};//* Function to perform desired operationvoid 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 reversevoid print(Node* head){ if (head == NULL) return; print(head->prev); cout << head->data; // Print linked list in reverse order}// Driver codeint 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 Classclass Node { public int data; public Node prev; public Node(int n) { data = n; prev = null; }}//* Function to perform desired operationclass 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 |
Output
Factorial of 100 is : 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
Time Complexity : O(N²)
Space Complexity : O(digits in factorial)
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