# Sum of the products of all possible Subsets

Last Updated : 12 Sep, 2023

Given an array of n non-negative integers. The task is to find the sum of the product of elements of all the possible subsets. It may be assumed that the numbers in subsets are small and computing product doesn’t cause arithmetic overflow.

Example :

Input : arr[] = {1, 2, 3}
Output : 23
Possible Subset are: 1, 2, 3, {1, 2}, {1, 3},
{2, 3}, {1, 2, 3}
Products of elements in above subsets :
1, 2, 3, 2, 3, 6, 6
Sum of all products = 1 + 2 + 3 + 2 + 3 + 6 + 6
= 23

Naive Approach: Simple approach is to generate all possible subset one by one and calculate sum of all elements. Time Complexity of this approach is exponential as there are total 2n – 1 subsets.

An Efficient approach is to generalize the whole problem into some pattern. Suppose we have two numbers a and b. We can write all possible subset products as:-

= a + b + ab
= a(1+b) + b + 1 - 1
= a(1+b) + (1+b) - 1
= (a + 1) * (b + 1) - 1
= (1+a) * (1 + b) - 1

Now take three numbers a, b, c:-

= a + b + c + ab + bc + ca + abc
= a + ac + b + bc + ab + abc + c + 1 - 1
= a(1+c) + b(1+c) + ab(1+c) + c + 1 - 1
= (a + b + ab + 1)(1+c) - 1
= (1+a) * (1+b) * (1+c) - 1

The above pattern can be generalized for n numbers.

Below is the implementation of above idea:

## C++

 // C++ program to find sum of product of // all subsets. #include using namespace std;    // Returns sum of products of all subsets // of arr[0..n-1] int productOfSubsetSums(int arr[], int n) {     int ans = 1;     for (int i = 0; i < n; ++i )         ans = ans * (arr[i] + 1);     return ans-1; }    // Driver code int main() {     int arr[] = {1, 2, 3, 4};     int n = sizeof(arr)/sizeof arr[0];     cout << productOfSubsetSums(arr, n);     return 0; }

## Java

 // Java program to find sum of product of // all subsets.    public class Subset {     // Returns sum of products of all subsets     // of arr[0..n-1]     static int productOfSubsetSums(int arr[], int n)     {         int ans = 1;         for (int i = 0; i < n; ++i )             ans = ans * (arr[i] + 1);         return ans-1;     }            public static void main (String[] args)     {         int arr[] = {1, 2, 3, 4};         int n = arr.length;         System.out.println(productOfSubsetSums(arr, n));     } }    // This code is contributed by Saket Kumar

## Python3

 # Python3 program to # find sum of product of # all subsets.    # Returns sum of products # of all subsets # of arr[0..n-1] def productOfSubsetSums(arr, n):     ans = 1;     for i in range(0,n):         ans = ans * (arr[i] + 1)     return ans-1    # Driver code arr = [1, 2, 3, 4] n = len(arr)    print (productOfSubsetSums(arr, n))        # This code is contributed # by Shreyanshi Arun.

## C#

 // C# program to find sum of  // product of all subsets. using System;    public class Subset {            // Returns sum of products of all      // subsets of arr[0..n-1]     static int productOfSubsetSums(int []arr, int n)     {         int ans = 1;         for (int i = 0; i < n; ++i )             ans = ans * (arr[i] + 1);         return ans-1;     }            // Driver Code     public static void Main ()     {         int []arr = {1, 2, 3, 4};         int n = arr.Length;         Console.Write(productOfSubsetSums(arr, n));     } }    // This code is contributed by Nitin Mittal.



## Javascript



Output

119

Time Complexity: O(n)
Auxiliary Space: O(1)

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