Related Articles

# Product of values of all possible non-empty subsets of given Array

• Last Updated : 01 Apr, 2021

Given an array of size N. The task is to find the product of values of all possible non-empty subsets of the given array.
Examples:

Input: N = 2, arr[] = {3, 7}
Output: 441
All non empty subsets are:

3, 7
Product = 3 * 7 * 3 * 7 = 441
Input: N = 1, arr[] = {4}
Output:

Approach: On careful observation it can be deduced that number of occurrences of every element across all subsets is 2N-1. Therefore, in the final product, every element of the array will be multiplied to itself by 2N-1 times.

`Product = a2N-1*a2N-1********a[N-1]2N-1`

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to find product of all elements``// in all subsets``int` `product(``int` `a[], ``int` `n)``{``    ``int` `ans = 1;``    ``int` `val = ``pow``(2, n - 1);``    ` `    ``for` `(``int` `i = 0; i < n; i++) {``        ``ans *= ``pow``(a[i], val);``    ``}``    ` `    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``int` `n = 2;``    ``int` `a[] = { 3, 7 };``    ` `    ``cout << product(a, n);``    ` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ` `    ``// Function to find product of all elements``    ``// in all subsets``    ``static` `int` `product(``int` `a[], ``int` `n)``    ``{``        ``int` `ans = ``1``;``        ``int` `val = (``int``)Math.pow(``2``, n - ``1``);``        ` `        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``ans *= (``int``)Math.pow(a[i], val);``        ``}``        ``return` `ans;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `n = ``2``;``        ``int` `a[] = { ``3``, ``7` `};``        ` `        ``System.out.println(product(a, n));``    ``}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach` `# Function to find product of``# all elements in all subsets``def` `product(a, n):``    ``ans ``=` `1``    ``val ``=` `pow``(``2``, n ``-` `1``)`  `    ``for` `i ``in` `range``(n):``        ``ans ``*``=` `pow``(a[i], val)` `    ``return` `ans` `# Driver Code``n ``=` `2``a ``=` `[``3``, ``7``]` `print``(product(a, n))` `# This code is contributed by mohit kumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `    ``// Function to find product of all elements``    ``// in all subsets``    ``static` `int` `product(``int` `[]a, ``int` `n)``    ``{``        ``int` `ans = 1;``        ``int` `val = (``int``)Math.Pow(2, n - 1);``        ` `        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``ans *= (``int``)Math.Pow(a[i], val);``        ``}``        ``return` `ans;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `n = 2;``        ``int` `[]a = { 3, 7 };``        ` `        ``Console.WriteLine(product(a, n));``    ``}``}` `// This code is contributed by anuj_67..`

## Javascript

 ``
Output:
`441`

Time Complexity: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up