Given an array with N positive integers. Find the GCD of factorials of all elements of array.
Examples:
Input : arr[] = {3, 4, 8, 6}
Output : 6
Input : arr[] = {13, 24, 8, 5}
Output : 120
Approach: To find the GCD of factorial of all elements, first of all, calculate the factorial of all elements and then find out their GCD. But this seems to be a very lengthy process. GCD of two numbers is the greatest number that divides both of the numbers. Hence, GCD of the factorial of two numbers is the value of the factorial of the smallest number itself.
For example, GCD of 3! (6) and 5! (120) is 3! (i.e. 6) itself.
Hence to find the GCD of factorial of all elements of the given array, find the smallest element and then print its factorial that will be our required answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Implementation of factorial functionint factorial(int n){ return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;}// Function to find GCD of factorial of // elements from arrayint gcdOfFactorial(int arr[], int n){ // find the minimum element of array int minm = arr[0]; for (int i = 1; i < n; i++) minm = minm > arr[i] ? arr[i] : minm; // return the factorial of minimum element return factorial(minm);}// Driver Codeint main(){ int arr[] = { 9, 12, 122, 34, 15 }; int n = sizeof(arr) / sizeof(arr[0]); cout << gcdOfFactorial(arr, n); return 0;} |
Java
// Java implementation of the above approachclass GFG{ // Implementation of factorial functionstatic int factorial(int n){ return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;}// Function to find GCD of factorial of // elements from arraystatic int gcdOfFactorial(int []arr, int n){ // find the minimum element of array int minm = arr[0]; for (int i = 1; i < n; i++) minm = minm > arr[i] ? arr[i] : minm; // return the factorial of minimum element return factorial(minm);}// Driver Codepublic static void main (String[] args) { int []arr = { 9, 12, 122, 34, 15 }; int n = arr.length; System.out.println(gcdOfFactorial(arr, n));}}// This code is contributed by mits |
Python3
# Implementation of factorial functiondef factorial(n): if n == 1 or n == 0: return 1 else: return factorial(n - 1) * n# Function to find GCD of factorial # of elements from arraydef gcdOfFactorial(arr, n): # find the minimum element # of array minm = arr[0] for i in range(1, n): if minm > arr[i]: minm = arr[i] else: arr[i] = minm # return the factorial of # minimum element return factorial(minm)# Driver Codearr = [9, 12, 122, 34, 15 ]n = len(arr)print(gcdOfFactorial(arr, n))# This code is contributed # by mohit kumar |
C#
// C# implementation of the above approachusing System;class GFG{ // Implementation of factorial functionstatic int factorial(int n){ return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;}// Function to find GCD of factorial of // elements from arraystatic int gcdOfFactorial(int []arr, int n){ // find the minimum element of array int minm = arr[0]; for (int i = 1; i < n; i++) minm = minm > arr[i] ? arr[i] : minm; // return the factorial of minimum element return factorial(minm);}// Driver Codestatic void Main(){ int []arr = { 9, 12, 122, 34, 15 }; int n = arr.Length; Console.WriteLine(gcdOfFactorial(arr, n));}}// This code is contributed by mits |
PHP
<?php// PHP implementation of the above approach // Implementation of factorial function function factorial($n) { return ($n == 1 || $n == 0) ? 1 : factorial($n - 1) * $n; } // Function to find GCD of factorial of // elements from array function gcdOfFactorial($arr, $n) { // find the minimum element of array $minm = $arr[0]; for ($i = 1; $i < $n; $i++) $minm = $minm > $arr[$i] ? $arr[$i] : $minm; // return the factorial of minimum element return factorial($minm); } // Driver Code $arr = array( 9, 12, 122, 34, 15 ); $n = count($arr); echo gcdOfFactorial($arr, $n);// This code is contributed by Srathore?> |
Javascript
<script>// JavaScript implementation of the above approach // Implementation of factorial function function factorial(n) { return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n; } // Function to find GCD of factorial of // elements from array function gcdOfFactorial(arr , n) { // find the minimum element of array var minm = arr[0]; for (i = 1; i < n; i++) minm = minm > arr[i] ? arr[i] : minm; // return the factorial of minimum element return factorial(minm); } // Driver Code var arr = [ 9, 12, 122, 34, 15 ]; var n = arr.length; document.write(gcdOfFactorial(arr, n));// This code is contributed by todaysgaurav </script> |
Output:
362880
Time Complexity: O(n)
Auxiliary Space: O(1)
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