Given a large number N, the task is to find the factorial of N using recursion.
Factorial of a non-negative integer is the multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is 720.
Examples:
Input : N = 100
Output : 933262154439441526816992388562667004-907159682643816214685929638952175999-932299156089414639761565182862536979-208272237582511852109168640000000000-00000000000000Input : N = 50
Output : 3041409320171337804361260816606476884-4377641568960512000000000000
Iterative Approach: The iterative approach is discussed in Set 1 of this article. Here, we have discussed the recursive approach.
Recursive Approach: To solve this problem recursively, the algorithm changes in the way that calls the same function recursively and multiplies the result by the number n. Follow the steps below to solve the problem:
- If n is less than equal to 2, then multiply n by 1 and store the result in a vector.
- Otherwise, call the function multiply(n, factorialRecursiveAlgorithm(n – 1)) to find the answer.
Below is the implementation of the above approach.
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// MUltiply the number x with the number// represented by res arrayvector<int> multiply(long int n, vector<int> digits){ // Initialize carry long int carry = 0; // One by one multiply n with // individual digits of res[] for (long int i = 0; i < digits.size(); i++) { long int result = digits[i] * n + carry; // Store last digit of 'prod' in res[] digits[i] = result % 10; // Put rest in carry carry = result / 10; } // Put carry in res and increase result size while (carry) { digits.push_back(carry % 10); carry = carry / 10; } return digits;}// Function to recursively calculate the// factorial of a large numbervector<int> factorialRecursiveAlgorithm( long int n){ if (n <= 2) { return multiply(n, { 1 }); } return multiply( n, factorialRecursiveAlgorithm(n - 1));}// Driver Codeint main(){ long int n = 50; vector<int> result = factorialRecursiveAlgorithm(n); for (long int i = result.size() - 1; i >= 0; i--) { cout << result[i]; } cout << "\n"; return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// MUltiply the number x with the number// represented by res arraystatic Integer []multiply(int n, Integer []digits){ // Initialize carry int carry = 0; // One by one multiply n with // individual digits of res[] for (int i = 0; i < digits.length; i++) { int result = digits[i] * n + carry; // Store last digit of 'prod' in res[] digits[i] = result % 10; // Put rest in carry carry = result / 10; } // Put carry in res and increase result size LinkedList<Integer> v = new LinkedList<Integer>(); v.addAll(Arrays.asList(digits)); while (carry>0) { v.add(new Integer(carry % 10)); carry = carry / 10; } return v.stream().toArray(Integer[] ::new);}// Function to recursively calculate the// factorial of a large numberstatic Integer []factorialRecursiveAlgorithm( int n){ if (n <= 2) { return multiply(n, new Integer[]{ 1 }); } return multiply( n, factorialRecursiveAlgorithm(n - 1));}// Driver Codepublic static void main(String[] args){ int n = 50; Integer []result = factorialRecursiveAlgorithm(n); for (int i = result.length - 1; i >= 0; i--) { System.out.print(result[i]); } System.out.print("\n");}}// This code is contributed by 29AjayKumar |
C#
// C# program for the above approachusing System;using System.Collections.Generic;using System.Linq;class GFG{ // MUltiply the number x with the number // represented by res array static int[] multiply(int n, int[] digits) { // Initialize carry int carry = 0; // One by one multiply n with // individual digits of res[] for (int i = 0; i < digits.Length; i++) { int result = digits[i] * n + carry; // Store last digit of 'prod' in res[] digits[i] = result % 10; // Put rest in carry carry = result / 10; } // Put carry in res and increase result size LinkedList<int> v = new LinkedList<int>(); foreach (int i in digits) { v.AddLast(i); } while (carry > 0) { v.AddLast((int)(carry % 10)); carry = carry / 10; } return v.ToArray(); } // Function to recursively calculate the // factorial of a large number static int[] factorialRecursiveAlgorithm( int n) { if (n <= 2) { return multiply(n, new int[] { 1 }); } return multiply( n, factorialRecursiveAlgorithm(n - 1)); } // Driver Code public static void Main() { int n = 50; int[] result = factorialRecursiveAlgorithm(n); for (int i = result.Length - 1; i >= 0; i--) { Console.Write(result[i]); } Console.Write("\n"); }}// This code is contributed by gfgking |
Python3
# Python 3 program for the above approach# MUltiply the number x with the number# represented by res arraydef multiply(n, digits): # Initialize carry carry = 0 # One by one multiply n with # individual digits of res[] for i in range(len(digits)): result = digits[i] * n + carry # Store last digit of 'prod' in res[] digits[i] = result % 10 # Put rest in carry carry = result // 10 # Put carry in res and increase result size while (carry): digits.append(carry % 10) carry = carry // 10 return digits# Function to recursively calculate the# factorial of a large numberdef factorialRecursiveAlgorithm(n): if (n <= 2): return multiply(n, [1]) return multiply( n, factorialRecursiveAlgorithm(n - 1))# Driver Codeif __name__ == "__main__": n = 50 result = factorialRecursiveAlgorithm(n) for i in range(len(result) - 1, -1, -1): print(result[i], end="") |
Javascript
<script>// javascript program for the above approach// MUltiply the number x with the number// represented by res arrayfunction multiply(n, digits){ // Initialize carry var carry = 0; // One by one multiply n with // individual digits of res for (var i = 0; i < digits.length; i++) { var result = digits[i] * n + carry; // Store last digit of 'prod' in res digits[i] = result % 10; // Put rest in carry carry = parseInt(result / 10); } // Put carry in res and increase result size while (carry>0) { digits.push(carry % 10); carry = parseInt(carry / 10); } return digits;}// Function to recursively calculate the// factorial of a large numberfunction factorialRecursiveAlgorithm( n){ if (n <= 2) { return multiply(n, [ 1 ]); } return multiply( n, factorialRecursiveAlgorithm(n - 1));}// Driver Code var n = 50; var result = factorialRecursiveAlgorithm(n); for (var i = result.length - 1; i >= 0; i--) { document.write(result[i]); } document.write("<br>");// This code is contributed by shikhasingrajput </script> |
Output
30414093201713378043612608166064768844377641568960512000000000000
Time Complexity: O(n*log(n))
Auxiliary Space: O(K), where K is the maximum number of digits in the output
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