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++
#include <bits/stdc++.h>
using namespace std;
int factorial( int n)
{
return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
}
int gcdOfFactorial( int arr[], int n)
{
int minm = arr[0];
for ( int i = 1; i < n; i++)
minm = minm > arr[i] ? arr[i] : minm;
return factorial(minm);
}
int main()
{
int arr[] = { 9, 12, 122, 34, 15 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << gcdOfFactorial(arr, n);
return 0;
}
|
Java
class GFG
{
static int factorial( int n)
{
return (n == 1 || n == 0 ) ? 1 : factorial(n - 1 ) * n;
}
static int gcdOfFactorial( int []arr, int n)
{
int minm = arr[ 0 ];
for ( int i = 1 ; i < n; i++)
minm = minm > arr[i] ? arr[i] : minm;
return factorial(minm);
}
public static void main (String[] args)
{
int []arr = { 9 , 12 , 122 , 34 , 15 };
int n = arr.length;
System.out.println(gcdOfFactorial(arr, n));
}
}
|
Python3
def factorial(n):
if n = = 1 or n = = 0 :
return 1
else :
return factorial(n - 1 ) * n
def gcdOfFactorial(arr, n):
minm = arr[ 0 ]
for i in range ( 1 , n):
if minm > arr[i]:
minm = arr[i]
else :
arr[i] = minm
return factorial(minm)
arr = [ 9 , 12 , 122 , 34 , 15 ]
n = len (arr)
print (gcdOfFactorial(arr, n))
|
C#
using System;
class GFG
{
static int factorial( int n)
{
return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
}
static int gcdOfFactorial( int []arr, int n)
{
int minm = arr[0];
for ( int i = 1; i < n; i++)
minm = minm > arr[i] ? arr[i] : minm;
return factorial(minm);
}
static void Main()
{
int []arr = { 9, 12, 122, 34, 15 };
int n = arr.Length;
Console.WriteLine(gcdOfFactorial(arr, n));
}
}
|
PHP
<?php
function factorial( $n )
{
return ( $n == 1 || $n == 0) ? 1 :
factorial( $n - 1) * $n ;
}
function gcdOfFactorial( $arr , $n )
{
$minm = $arr [0];
for ( $i = 1; $i < $n ; $i ++)
$minm = $minm > $arr [ $i ] ?
$arr [ $i ] : $minm ;
return factorial( $minm );
}
$arr = array ( 9, 12, 122, 34, 15 );
$n = count ( $arr );
echo gcdOfFactorial( $arr , $n );
?>
|
Javascript
<script>
function factorial(n) {
return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
}
function gcdOfFactorial(arr , n) {
var minm = arr[0];
for (i = 1; i < n; i++)
minm = minm > arr[i] ? arr[i] : minm;
return factorial(minm);
}
var arr = [ 9, 12, 122, 34, 15 ];
var n = arr.length;
document.write(gcdOfFactorial(arr, n));
</script>
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Time Complexity: O(n)
Auxiliary Space: O(1)