Product of first N factorials

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 *.

Steps to solve the problem:

• Define a function named mulmod() that takes three arguments a, b, and mod.
•  Inside mulmod(), initialize res to 0.
•  Compute a = a % mod.
• While b > 0, perform the following operations:
  • If b % 2 == 1, add a to res modulo mod.
  • Multiply a by 2 and compute a = a % mod.
• Divide b by 2 and discard the remainder. Return res % mod.
•  Define a function named findProduct() that takes a single argument N.
•  Inside findProduct(), initialize product and fact to 1.
•  Set MOD = 1e9 + 7.
•  For i in the range 1 to N, perform the following operations:
  • Compute fact as the product of fact and i modulo MOD, i.e., fact = mulmod(fact, i, MOD)
  • Compute product as the product of product and fact modulo MOD, i.e., product = mulmod(product, fact, MOD).
  • If product is zero, return 0.
•  Return product

Below is the implementation of the above approach. 

C++

// CPP Program to find the
// product of first N factorials
#include <bits/stdc++.h>

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

<?php
// PHP Program to find the
// product of first N factorials

// To compute (a * b) % MOD
function 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
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

<script>
    // Javascript Program to find the
    // product of first N factorials

    // To compute (a * b) % MOD
    function mulmod(a, b, mod)
    {
        let 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 = parseInt(b / 2, 10);
        }

        // 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
        let product = 1, fact = 1;
        let MOD = (1e9 + 7);
        for (let 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;
    }

    let N = 3;
    document.write(findProduct(N) + "</br>");

    N = 5;
    document.write(findProduct(N));

</script>

Output:

12
34560

Time Complexity: O(N * logN), where O(log N) is the time complexity of modular multiplication.
Auxiliary Space: O(1) because it is using constant space for variables