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# GCD of factorials of two numbers

Given two numbers m and n. Find the GCD of  their factorial.
Examples :

Input : n = 3, m = 4
Output : 6
Explanation:
Factorial of n = 1 * 2 * 3 = 6
Factorial of m = 1 * 2 * 3 * 4 = 24
GCD(6, 24) = 6.

Input : n = 9, m = 5
Output : 20
Explanation:
Factorial of n = 1 * 2 * 3 *4 * 5 * 6 * 7 * 8 * 9 = 362880
Factorial of m = 1 * 2 * 3 * 4 * 5 = 120
GCD(362880, 120) = 120

A simple solution is to find factorials of both numbers. Then find GCD of the computed factorials.
An efficient solution is based on the fact that GCD of two factorials is equal to smaller factorial (note that factorials have all terms common).
Below is the implementation of above approach.

## C++

 `// CPP program to find GCD of factorial of two``// numbers.``#include ``using` `namespace` `std;` `int` `factorial(``int` `x)``{``    ``if` `(x <= 1)``        ``return` `1;``    ``int` `res = 2;``    ``for` `(``int` `i = 3; i <= x; i++)``        ``res = res * i;``    ``return` `res;``}` `int` `gcdOfFactorial(``int` `m, ``int` `n)``{``    ``return` `factorial(min(m, n));``}` `int` `main()``{``    ``int` `m = 5, n = 9;``    ``cout << gcdOfFactorial(m, n);``    ``return` `0;``}`

## Java

 `// Java program to find GCD of factorial``// of two numbers.``public` `class` `FactorialGCD{``    ` `static` `int` `factorial(``int` `x)``{``    ``if` `(x <= ``1``)``        ``return` `1``;``    ``int` `res = ``2``;``    ``for` `(``int` `i = ``3``; i <= x; i++)``        ``res = res * i;``    ``return` `res;``}` `static` `int` `gcdOfFactorial(``int` `m, ``int` `n)``{``    ``int` `min = m < n ? m : n;``    ``return` `factorial(min);``}` `    ``/* Driver program to test above functions */``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `m = ``5``, n = ``9``;``        ` `        ``System.out.println(gcdOfFactorial(m, n));``    ``}``}` `// This code is contributed by Prerna Saini`

## Python

 `# Python code to find GCD of factorials of``# two numbers.``import` `math` `def` `gcdOfFactorial(m, n) :``    ``return` `math.factorial(``min``(m, n))` `# Driver code``m ``=` `5``n ``=` `9``print``(gcdOfFactorial(m, n))`

## C#

 `// C# program to find GCD of factorial``// of two numbers.``using` `System;` `public` `class` `GFG{``     ` `    ``static` `int` `factorial(``int` `x)``    ``{``        ``if` `(x <= 1)``            ``return` `1;``            ` `        ``int` `res = 2;``        ` `        ``for` `(``int` `i = 3; i <= x; i++)``            ``res = res * i;``            ` `        ``return` `res;``    ``}``     ` `    ``static` `int` `gcdOfFactorial(``int` `m, ``int` `n)``    ``{``        ``int` `min = m < n ? m : n;``        ``return` `factorial(min);``    ``}`` ` `    ``/* Driver program to test above functions */``    ``public` `static` `void` `Main()``    ``{``        ` `        ``int` `m = 5, n = 9;``         ` `        ``Console.WriteLine(gcdOfFactorial(m, n));``    ``}``}`` ` `// This code is contributed by Anant Agarwal.`

## PHP

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## Javascript

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Output

`120`

Time complexity: O(min(m,n))
Auxiliary space: O(1)

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