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 arri 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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

4

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