Product of all Subsets of a set formed by first N natural numbers

Given a number N, the task is to find the product of all the elements from all possible subsets of a set formed by first N natural numbers.
Examples:

Input: N = 2 
Output: 4 
Possible subsets are {{1}, {2}, {1, 2}}. 
Product of elements in subsets = {1} * {2} * {1 * 2} = 4
Input: N = 3 
Output: 1296 
Possible subsets are {{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} 
Product of elements in subsets = 1 * 2 * 3 * (1 * 2) * (1 * 3) * (2 * 3) * (1 * 2 * 3) = 1296 

Naive Approach: A simple solution is to generate all subsets of first N natural number. Then for every subset, compute its product and finally return overall product of each subset.
Efficient Approach:

• It can be observed that each element of the original array appears in 2^{(N – 1)} times in all subsets.
• Therefore contribution of any element arr_i in the final answer will be 
  

i * 2(N – 1)

• So, the Sum of cubes of all Subsets will be 
  

12N-1 * 22N-1 * 32N-1......N2N-1

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the product of all elements
// in all subsets in natural numbers from 1 to N
int product(int N)
{
    int ans = 1;
    int val = pow(2, N - 1);

    for (int i = 1; i <= N; i++) {
        ans *= pow(i, val);
    }

    return ans;
}

// Driver Code
int main()
{
    int N = 2;

    cout << product(N);

    return 0;
}

Java

// Java implementation of the approach
class GFG {

    // Function to find the product of all elements
    // in all subsets in natural numbers from 1 to N
    static int product(int N)
    {
        int ans = 1;
        int val = (int)Math.pow(2, N - 1);

        for (int i = 1; i <= N; i++) {
            ans *= (int)Math.pow(i, val);
        }

        return ans;
    }

    // Driver Code
    public static void main (String[] args)
    {
        int N = 2;

        System.out.println(product(N));
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach

# Function to find the product of all elements
# in all subsets in natural numbers from 1 to N
def product(N) :
    ans = 1;
    val = 2 **(N - 1);

    for i in range(1, N + 1) :
        ans *= (i**val);

    return ans;


# Driver Code
if __name__ == "__main__" :

    N = 2;

    print(product(N));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG {

    // Function to find the product of all elements
    // in all subsets in natural numbers from 1 to N
    static int product(int N)
    {
        int ans = 1;
        int val = (int)Math.Pow(2, N - 1);

        for (int i = 1; i <= N; i++) {
            ans *= (int)Math.Pow(i, val);
        }

        return ans;
    }

    // Driver Code
    public static void Main (string[] args)
    {
        int N = 2;

        Console.WriteLine(product(N));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>
// javascript implementation of the approach

// Function to find the product of all elements
// in all subsets in natural numbers from 1 to N
function product( N)
{
    let ans = 1;
    let val = Math.pow(2, N - 1);
    for (let i = 1; i <= N; i++)
    {
        ans *= Math.pow(i, val);
    }
    return ans;
}

// Driver Code

    let N = 2;
    document.write(product(N));

// This code is contributed by todaysgaurav

</script>

Output:

4

Time Complexity: O(N*logN)

Auxiliary Space: O(1)