# Sum of the sums of all possible subsets

Given an array a of size N. The task is to find the sum of the sums of all possible subsets.
Examples:

Input: a[] = {3, 7}
Output: 20
The subsets are: {3} {7} {3, 7}
{3, 7} = 10
{3} = 3
{7} = 7
10 + 3 + 7 = 20
Input: a[] = {10, 16, 14, 9}
Output: 392

Naive Approach: A naive approach is to find all the subsets using power set and then summate all the possible subsets to get the answer.

## C++

 `// C++ program to check if there is a subset` `// with sum divisible by m.` `#include ` `using` `namespace` `std;`   `int` `helper(``int` `N, ``int` `nums[], ``int` `sum, ``int` `idx)` `{` `    ``// if we reach last index` `    ``if` `(idx == N) {` `        ``// and if the sum mod m is zero` `        ``return` `sum;` `    ``}`   `    ``// 2 choices - to pick or to not pick` `    ``int` `picked = helper(N, nums, sum + nums[idx], idx + 1);` `    ``int` `notPicked = helper(N, nums, sum, idx + 1);`   `    ``return` `picked + notPicked;` `}`   `int` `sumOfSubset(``int` `arr[], ``int` `n)` `{` `    ``return` `helper(n, arr, 0, 0);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 3, 7 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``cout << sumOfSubset(arr, n);`   `    ``return` `0;` `}`

## Java

 `/*package whatever //do not write package name here */`   `import` `java.io.*;`   `class` `GFG {`   `    ``static` `int` `helper(``int` `N, ``int` `nums[], ``int` `sum, ``int` `idx)` `    ``{` `        ``// if we reach last index` `        ``if` `(idx == N) {` `            ``// and if the sum mod m is zero` `            ``return` `sum;` `        ``}`   `        ``// 2 choices - to pick or to not pick` `        ``int` `picked` `            ``= helper(N, nums, sum + nums[idx], idx + ``1``);` `        ``int` `notPicked = helper(N, nums, sum, idx + ``1``);`   `        ``return` `picked + notPicked;` `    ``}`   `    ``static` `int` `sumOfSubset(``int` `arr[], ``int` `n)` `    ``{` `        ``return` `helper(n, arr, ``0``, ``0``);` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``int` `arr[] = { ``3``, ``7` `};` `        ``int` `n = arr.length;`   `        ``System.out.println(sumOfSubset(arr, n));` `    ``}` `}`   `// This code is contributed by aadityaburujwale.`

## Python3

 `# Python program to check if there is a subset` `# with sum divisible by m.` `def` `helper(N, nums, ``sum``, idx):` `    ``# if we reach last index` `    ``if` `idx ``=``=` `N:` `        ``# and if the sum mod m is zero` `        ``return` `sum`   `    ``# 2 choices - to pick or to not pick` `    ``picked ``=` `helper(N, nums, ``sum` `+` `nums[idx], idx ``+` `1``)` `    ``not_picked ``=` `helper(N, nums, ``sum``, idx ``+` `1``)`   `    ``return` `picked ``+` `not_picked`   `def` `sum_of_subset(arr, n):` `    ``return` `helper(n, arr, ``0``, ``0``)`   `# Test the program` `arr ``=` `[``3``, ``7``]` `n ``=` `len``(arr)`   `print``(sum_of_subset(arr, n))`   `# This code is contributed by divyansh2212`

## C#

 `/*package whatever //do not write package name here */` `using` `System;`   `class` `GFG {`   `  ``static` `int` `helper(``int` `N, ``int``[] nums, ``int` `sum, ``int` `idx)` `  ``{` `    ``// if we reach last index` `    ``if` `(idx == N) {` `      ``// and if the sum mod m is zero` `      ``return` `sum;` `    ``}`   `    ``// 2 choices - to pick or to not pick` `    ``int` `picked` `      ``= helper(N, nums, sum + nums[idx], idx + 1);` `    ``int` `notPicked = helper(N, nums, sum, idx + 1);`   `    ``return` `picked + notPicked;` `  ``}`   `  ``static` `int` `sumOfSubset(``int``[] arr, ``int` `n)` `  ``{` `    ``return` `helper(n, arr, 0, 0);` `  ``}`   `  ``public` `static` `void` `Main()` `  ``{`   `    ``int``[] arr = { 3, 7 };` `    ``int` `n = arr.Length;`   `    ``Console.WriteLine(sumOfSubset(arr, n));` `  ``}` `}`   `// This code is contributed by Samim Hossain Mondal.`

## Javascript

 `// JS program to check if there is a subset` `// with sum divisible by m.` `function` `helper(N, nums, sum, idx)` `{` `    ``// if we reach last index` `    ``if` `(idx == N)` `    ``{` `    `  `        ``// and if the sum mod m is zero` `        ``return` `sum;` `    ``}`   `    ``// 2 choices - to pick or to not pick` `    ``let picked = helper(N, nums, sum + nums[idx], idx + 1);` `    ``let notPicked = helper(N, nums, sum, idx + 1);`   `    ``return` `picked + notPicked;` `}`   `function` `sumOfSubset(arr, n)` `{` `    ``return` `helper(n, arr, 0, 0);` `}`   `// Driver code` `let arr = [ 3, 7 ];` `let n = arr.length;`   `console.log(sumOfSubset(arr, n));`   `// This code is contributed by akashish__`

Output

`20`

Time Complexity: O(2N)

Space Complexity: O(N) because of Recursion Stack Space
Efficient Approach: An efficient approach is to solve the problem using observation. If we write all the subsequences, a common point of observation is that each number appears 2(N – 1) times in a subset and hence will lead to the 2(N-1) as the contribution to the sum. Iterate through the array and add (arr[i] * 2N-1) to the answer.
Below is the implementation of the above approach:

## C++

 `// C++ program to find the sum of` `// the addition of all possible subsets.` `#include ` `using` `namespace` `std;`   `// Function to find the sum` `// of sum of all the subset` `int` `sumOfSubset(``int` `a[], ``int` `n)` `{` `    ``int` `times = ``pow``(2, n - 1);`   `    ``int` `sum = 0;`   `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``sum = sum + (a[i] * times);` `    ``}`   `    ``return` `sum;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `a[] = { 3, 7 };` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);` `    ``cout << sumOfSubset(a, n);` `}`

## Java

 `// Java program to find the sum of` `// the addition of all possible subsets.` `class` `GFG` `{` `    `  `// Function to find the sum` `// of sum of all the subset` `static` `int` `sumOfSubset(``int` `[]a, ``int` `n)` `{` `    ``int` `times = (``int``)Math.pow(``2``, n - ``1``);`   `    ``int` `sum = ``0``;`   `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{` `        ``sum = sum + (a[i] * times);` `    ``}`   `    ``return` `sum;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `[]a = { ``3``, ``7` `};` `    ``int` `n = a.length;` `    ``System.out.println(sumOfSubset(a, n));` `}` `}`   `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to find the Sum of` `# the addition of all possible subsets.`   `# Function to find the sum` `# of sum of all the subset` `def` `SumOfSubset(a, n):`   `    ``times ``=` `pow``(``2``, n ``-` `1``)`   `    ``Sum` `=` `0`   `    ``for` `i ``in` `range``(n):` `        ``Sum` `=` `Sum` `+` `(a[i] ``*` `times)`   `    ``return` `Sum`   `# Driver Code` `a ``=` `[``3``, ``7``]` `n ``=` `len``(a)` `print``(SumOfSubset(a, n))`   `# This code is contributed by Mohit Kumar`

## C#

 `// C# program to find the sum of` `// the addition of all possible subsets.` `using` `System;`   `class` `GFG` `{` `    `  `// Function to find the sum` `// of sum of all the subset` `static` `int` `sumOfSubset(``int` `[]a, ``int` `n)` `{` `    ``int` `times = (``int``)Math.Pow(2, n - 1);`   `    ``int` `sum = 0;`   `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{` `        ``sum = sum + (a[i] * times);` `    ``}`   `    ``return` `sum;` `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `    ``int` `[]a = { 3, 7 };` `    ``int` `n = a.Length;` `    ``Console.Write(sumOfSubset(a, n));` `}` `}`   `// This code is contributed by Nidhi`

## Javascript

 ``

Output:

`20`

Time Complexity: O(N)
Space Complexity: O(1)
Note: If N is large, the answer can overflow, thereby use larger data-type.

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