Given a positive number N, the task is to calculate the factorial of N without using the multiplication operator.
Examples:
Input: N = 5
Output:
120
Explanation:
5*4*3*2*1=120Input: N = 7
Output:
5040
Observation:
A*B=A+A+A+A...B times. This observation can be used as follows: 5!=5*4*3*2*1 =(5+5+5+5)*3*2*1 =(20+20+20)*2*1 =60+60 =120
Approach 1: The problem can be solved using the concept of nested loops. Instead of using the multiplication operator, the answer can be manually calculated by using another loop. Follow the steps below to solve the problem:
- Initialize a variable ans to N.
- Iterate from N-1 to 1, using the variable i, and do the following:
- Initialize a variable sum to 0.
- Iterate from 0 to i-1, using the variable j, and add ans to sum
- Add sum to ans.
- Print ans.
Below is the implementation of the above approach
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate factorial of the number// without using multiplication operatorint factorialWithoutMul(int N){ // variable to store the final factorial int ans = N; // Outer loop for (int i = N - 1; i > 0; i--) { int sum = 0; // Inner loop for (int j = 0; j < i; j++) sum += ans; ans = sum; } return ans;}// Driver codeint main(){ // Input int N = 5; // Function calling cout << factorialWithoutMul(N) << endl; return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG { // Function to calculate factorial of the number // without using multiplication operator public static int factorialWithoutMul(int N) { // variable to store the final factorial int ans = N; // Outer loop for (int i = N - 1; i > 0; i--) { int sum = 0; // Inner loop for (int j = 0; j < i; j++) sum += ans; ans = sum; } return ans; } // Driver code public static void main(String[] args) { int N = 5; // Function calling System.out.println(factorialWithoutMul(N)); // This code is contributed by Potta Lokesh }} |
Python3
# Python3 program for the above approach# Function to calculate factorial of the number# without using multiplication operatordef factorialWithoutMul(N): # Variable to store the final factorial ans = N # Outer loop i = N - 1 while (i > 0): sum = 0 # Inner loop for j in range(i): sum += ans ans = sum i -= 1 return ans# Driver codeif __name__ == '__main__': # Input N = 5 # Function calling print(factorialWithoutMul(N))# This code is contributed by SURENDRA_GANGWAR |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG { // Function to calculate factorial of the number // without using multiplication operator static int factorialWithoutMul(int N) { // Variable to store the final factorial int ans = N; // Outer loop for (int i = N - 1; i > 0; i--) { int sum = 0; // Inner loop for (int j = 0; j < i; j++) sum += ans; ans = sum; } return ans; } // Driver code public static void Main() { // Input int N = 5; // Function calling Console.Write(factorialWithoutMul(N)); }}// This code is contributed by SURENDRA_GANGWAR |
Javascript
<script>// JavaScript program for the above approach// Function to calculate factorial of the number// without using multiplication operatorfunction factorialWithoutMul(N){ // variable to store the final factorial let ans = N; // Outer loop for (let i = N - 1; i > 0; i--) { let sum = 0; // Inner loop for (let j = 0; j < i; j++) sum += ans; ans = sum; } return ans;}// Driver code // Input let N = 5; // Function calling document.write(factorialWithoutMul(N)); // This code is contributed by Potta Lokesh </script> |
Output
120
Time Complexity: O(N2)
Auxiliary Space: O(1)
Approach 2: The problem can be solved by dividing with the reciprocal of the next number instead of multiplying it. In algebra, ab also means a / (1/b). We will be using the same concept to find the factorial of a number without using asterisk. Follow the steps below to solve the problem:
- Take a variable of integer type (here: n) which would store the value of which we’re finding the factorial.
- Initialize a variable (here: p) which would serve to be the factorial of n.
- Start iterating from n to 1. In each step, set the value of p to be the same as that of p divided by the reciprocal of iterator (here: i).
- Print p.
Below is the implementation of the above approach in java:
C++
// C++ code for the above approach#include <iostream>using namespace std;int factorial(int n){ // Function to find the factorial of (n) without // multiplying. int p = 1; for (int i = n; i >= 1; i--) { // Loop to calculate the factorial of (n). p = p / (1.0 / i); } // Returning the factorial of (n) stored in (p). return p;}int main(){ int n = 5; // Printing the factorial of (n). cout << factorial(n) << endl; return 0;}// This code is contributed by lokesh. |
Java
public class Factorial { int factorial(int n) { // Function to find the factorial of (n) without // multiplying. int p = 1; for (int i = n; i >= 1; i--) { // Loop to calculate the factorial of // (n). p = (int)(p / (1.0 / i)); } return p; // Returning the factorial of (n) stored // in (p). } public static void main(String[] Args) { Factorial fact = new Factorial(); // Creating an instance of // Factorial class. int n = 5; System.out.println(fact.factorial( n)); // Printing the factorial of (n). }} |
Python3
# Python3 code for the above approachdef factorial(n): # Function to find the factorial of (n) without multiplying. p = 1 for i in range(n, 0, -1): # Loop to calculate the factorial of (n). p = p / (1.0 / i) # Returning the factorial of (n) stored in (p). return p# Driver coden = 5# Printing the factorial of (n).print(factorial(n))# This code is contributed by phasing17. |
C#
// C# code for the above approachusing System;public class Factorial { int factorial(int n) { // Function to find the factorial of (n) without // multiplying. int p = 1; for (int i = n; i >= 1; i--) { // Loop to calculate the factorial of (n). p = (int)(p / (1.0 / i)); } // Returning the factorial of (n) stored in (p). return p; } static public void Main() { Factorial fact = new Factorial(); // Creating an instance of // Factorial class. int n = 5; Console.WriteLine(fact.factorial( n)); // Printing the factorial of (n). }}// This code is contributed by lokeshmvs21. |
Javascript
// JavaScript code for the above approachfunction factorial(n){ // Function to find the factorial of (n) without // multiplying. let p = 1; for (let i = n; i >= 1; i--) { // Loop to calculate the factorial of (n). p = p / (1.0 / i); } // Returning the factorial of (n) stored in (p). return p;} let n = 5; // Printing the factorial of (n). console.log(factorial(n))// This code is contributed by poojaagarwal2. |
Output
120
Time Complexity: O(N)
Auxiliary Space: O(1)
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