# Product of first N factorials

• Difficulty Level : Basic
• Last Updated : 05 May, 2021

Given a number N. Find the product of first N factorials modulo 1000000007.

Constraints: 1 ≤ N ≤ 1e6

Examples:

```Input : 3
Output : 12
Explanation: 1! * 2! * 3! = 12 mod (1e9 + 7) = 12

Input : 5
Output : 34560```

Prerequisites: Modular Multiplication
Approach: The basic idea behind solving this problem is to just consider the problem of overflow during the multiplication of such large numbers i.e. factorials. Hence, it needs to be addressed by multiplying recursively to overcome the difficulty of overflow. Moreover, we have to take modulus at every step while computing factorials iteratively and modular multiplication.

```facti = facti-1 * i
where facti is the factorial of ith number

prodi = prodi-1 * facti
where prodi is the product of first i factorials```

To find the product of two large numbers under modulo, we use the same approach as exponentiation under modulo… In the multiplication function, we use + instead of *.
Below is the implementation of the above approach.

## C++

 `// CPP Program to find the``// product of first N factorials``#include ` `using` `namespace` `std;` `// To compute (a * b) % MOD``long` `long` `int` `mulmod(``long` `long` `int` `a, ``long` `long` `int` `b,``                                    ``long` `long` `int` `mod)``{``    ``long` `long` `int` `res = 0; ``// Initialize result``    ``a = a % mod;``    ``while` `(b > 0) {` `        ``// If b is odd, add 'a' to result``        ``if` `(b % 2 == 1)``            ``res = (res + a) % mod;` `        ``// Multiply 'a' with 2``        ``a = (a * 2) % mod;` `        ``// Divide b by 2``        ``b /= 2;``    ``}` `    ``// Return result``    ``return` `res % mod;``}` `// This function computes factorials and``// product by using above function i.e.``// modular multiplication``long` `long` `int` `findProduct(``long` `long` `int` `N)``{``    ``// Initialize product and fact with 1``    ``long` `long` `int` `product = 1, fact = 1;``    ``long` `long` `int` `MOD = 1e9 + 7;``    ``for` `(``int` `i = 1; i <= N; i++) {` `        ``// ith factorial``        ``fact = mulmod(fact, i, MOD);` `        ``// product of first i factorials``        ``product = mulmod(product, fact, MOD);` `        ``// If at any iteration, product becomes``        ``// divisible by MOD, simply return 0;``        ``if` `(product == 0)``            ``return` `0;``    ``}``    ``return` `product;``}` `// Driver Code to Test above functions``int` `main()``{``    ``long` `long` `int` `N = 3;``    ``cout << findProduct(N) << endl;` `    ``N = 5;``    ``cout << findProduct(N) << endl;` `    ``return` `0;``}`

## Java

 `// Java Program to find the``// product of first N factorials` `class` `GFG{``// To compute (a * b) % MOD``static` `double` `mulmod(``long` `a, ``long` `b,``                                    ``long` `mod)``{``    ``long` `res = ``0``; ``// Initialize result``    ``a = a % mod;``    ``while` `(b > ``0``) {` `        ``// If b is odd, add 'a' to result``        ``if` `(b % ``2` `== ``1``)``            ``res = (res + a) % mod;` `        ``// Multiply 'a' with 2``        ``a = (a * ``2``) % mod;` `        ``// Divide b by 2``        ``b /= ``2``;``    ``}` `    ``// Return result``    ``return` `res % mod;``}` `// This function computes factorials and``// product by using above function i.e.``// modular multiplication``static` `long` `findProduct(``long` `N)``{``    ``// Initialize product and fact with 1``    ``long` `product = ``1``, fact = ``1``;``    ``long` `MOD = (``long``)(1e9 + ``7``);``    ``for` `(``int` `i = ``1``; i <= N; i++) {` `        ``// ith factorial``        ``fact = (``long``)mulmod(fact, i, MOD);` `        ``// product of first i factorials``        ``product = (``long``)mulmod(product, fact, MOD);` `        ``// If at any iteration, product becomes``        ``// divisible by MOD, simply return 0;``        ``if` `(product == ``0``)``            ``return` `0``;``    ``}``    ``return` `product;``}` `// Driver Code to Test above functions``public` `static` `void` `main(String[] args)``{``    ``long` `N = ``3``;``    ``System.out.println(findProduct(N));` `    ``N = ``5``;``    ``System.out.println(findProduct(N));` `}``}``// this Code is contributed by mits`

## Python3

 `# Python Program to find the``# product of first N factorials` `# To compute (a * b) % MOD``def` `mulmod(a, b, mod):``    ``res ``=` `0` `# Initialize result``    ``a ``=` `a ``%` `mod``    ``while` `(b > ``0``):` `        ``# If b is odd, add 'a' to result``        ``if` `(b ``%` `2` `=``=` `1``):``            ``res ``=` `(res ``+` `a) ``%` `mod` `        ``# Multiply 'a' with 2``        ``a ``=` `(a ``*` `2``) ``%` `mod` `        ``# Divide b by 2``        ``b ``/``/``=` `2` `    ``# Return result``    ``return` `res ``%` `mod` `# This function computes factorials and``# product by using above function i.e.``# modular multiplication``def` `findProduct(N):``    ``# Initialize product and fact with 1``    ``product ``=` `1``; fact ``=` `1``    ``MOD ``=` `1e9` `+` `7``    ``for` `i ``in` `range``(``1``, N``+``1``):` `        ``# ith factorial``        ``fact ``=` `mulmod(fact, i, MOD)` `        ``# product of first i factorials``        ``product ``=` `mulmod(product, fact, MOD)` `        ``# If at any iteration, product becomes``        ``# divisible by MOD, simply return 0``        ``if` `not` `product:``            ``return` `0``    ``return` `int``(product)` `# Driver Code to Test above functions``N ``=` `3``print``(findProduct(N))``N ``=` `5``print``(findProduct(N))` `# This code is contributed by Ansu Kumari`

## C#

 `// C#  Program to find the``// product of first N factorials` `using` `System;` `public` `class` `GFG{``    ``// To compute (a * b) % MOD``static` `double` `mulmod(``long` `a, ``long` `b,``                                    ``long` `mod)``{``    ``long` `res = 0; ``// Initialize result``    ``a = a % mod;``    ``while` `(b > 0) {` `        ``// If b is odd, add 'a' to result``        ``if` `(b % 2 == 1)``            ``res = (res + a) % mod;` `        ``// Multiply 'a' with 2``        ``a = (a * 2) % mod;` `        ``// Divide b by 2``        ``b /= 2;``    ``}` `    ``// Return result``    ``return` `res % mod;``}` `// This function computes factorials and``// product by using above function i.e.``// modular multiplication``static` `long` `findProduct(``long` `N)``{``    ``// Initialize product and fact with 1``    ``long` `product = 1, fact = 1;``    ``long` `MOD = (``long``)(1e9 + 7);``    ``for` `(``int` `i = 1; i <= N; i++) {` `        ``// ith factorial``        ``fact = (``long``)mulmod(fact, i, MOD);` `        ``// product of first i factorials``        ``product = (``long``)mulmod(product, fact, MOD);` `        ``// If at any iteration, product becomes``        ``// divisible by MOD, simply return 0;``        ``if` `(product == 0)``            ``return` `0;``    ``}``    ``return` `product;``}` `// Driver Code to Test above functions``    ``static` `public` `void` `Main (){``        ``long` `N = 3;``        ``Console.WriteLine(findProduct(N));``        ``N = 5;``        ``Console.WriteLine(findProduct(N));` `}``}``//This Code is contributed by ajit.`

## PHP

 ` 0)``    ``{` `        ``// If b is odd, add 'a' to result``        ``if` `(``\$b` `% 2 == 1)``            ``\$res` `= (``\$res` `+ ``\$a``) % ``\$mod``;` `        ``// Multiply 'a' with 2``        ``\$a` `= (``\$a` `* 2) % ``\$mod``;` `        ``// Divide b by 2``        ``\$b` `/= 2;``    ``}` `    ``// Return result``    ``return` `\$res` `% ``\$mod``;``}` `// This function computes factorials and``// product by using above function i.e.``// modular multiplication``function` `findProduct(``\$N``)``{``    ``// Initialize product and fact with 1``    ``\$product` `= 1;``    ``\$fact` `= 1;``    ``\$MOD` `= 1000000000;``    ``for` `(``\$i` `= 1; ``\$i` `<= ``\$N``; ``\$i``++)``    ``{` `        ``// ith factorial``        ``\$fact` `= mulmod(``\$fact``, ``\$i``, ``\$MOD``);` `        ``// product of first i factorials``        ``\$product` `= mulmod(``\$product``, ``\$fact``, ``\$MOD``);` `        ``// If at any iteration, product becomes``        ``// divisible by MOD, simply return 0;``        ``if` `(``\$product` `== 0)``            ``return` `0;``    ``}``    ``return` `\$product``;``}` `// Driver Code``\$N` `= 3;``echo` `findProduct(``\$N``),``"\n"``;` `\$N` `= 5;``echo` `findProduct(``\$N``),``"\n"``;` `// This code is contributed by ajit``?>`

## Javascript

 ``
Output:
```12
34560```

Time Complexity: O(N * logN), where O(log N) is the time complexity of modular multiplication.

My Personal Notes arrow_drop_up