# Sum of subsets of all the subsets of an array | O(3^N)

• Difficulty Level : Medium
• Last Updated : 10 Jun, 2022

Given an array arr[] of length N, the task is to find the overall sum of subsets of all the subsets of the array.
Examples:

Input: arr[] = {1, 1}
Output:
All possible subsets:
a) {} : 0
All the possible subsets of this subset
will be {}, Sum = 0
b) {1} : 1
All the possible subsets of this subset
will be {} and {1}, Sum = 0 + 1 = 1
c) {1} : 1
All the possible subsets of this subset
will be {} and {1}, Sum = 0 + 1 = 1
d) {1, 1} : 4
All the possible subsets of this subset
will be {}, {1}, {1} and {1, 1}, Sum = 0 + 1 + 1 + 2 = 4
Thus, ans = 0 + 1 + 1 + 4 = 6
Input: arr[] = {1, 4, 2, 12}
Output: 513

Approach: In this article, an approach with O(3N) time complexity to solve the given problem will be discussed.
First, generate all the possible subsets of the array. There will be 2N subsets in total. Then for each subset, find the sum of all of its subset.
Now, let’s understand the time-complexity of this solution.
There are NCk subsets of length K and time to find the subsets of an array of length K is 2K
Total time = (NC1 * 21) + (NC2 * 22) + … + (NCk * K) = 3K
Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach#include using namespace std; // Function to sum of all subsets of a// given arrayvoid subsetSum(vector& c, int i,               int& ans, int curr){    // Base case    if (i == c.size()) {        ans += curr;        return;    }     // Recursively calling subsetSum    subsetSum(c, i + 1, ans, curr + c[i]);    subsetSum(c, i + 1, ans, curr);} // Function to generate the subsetsvoid subsetGen(int* arr, int i, int n,               int& ans, vector& c){    // Base-case    if (i == n) {         // Finding the sum of all the subsets        // of the generated subset        subsetSum(c, 0, ans, 0);        return;    }     // Recursively accepting and rejecting    // the current number    subsetGen(arr, i + 1, n, ans, c);    c.push_back(arr[i]);    subsetGen(arr, i + 1, n, ans, c);    c.pop_back();} // Driver codeint main(){    int arr[] = { 1, 1 };    int n = sizeof(arr) / sizeof(int);     // To store the final ans    int ans = 0;    vector c;     subsetGen(arr, 0, n, ans, c);    cout << ans;     return 0;}

## Java

 // Java implementation of the approachimport java.util.*; class GFG{    static Vector c = new Vector<>();     // To store the final ans    static int ans = 0;     // Function to sum of all subsets of a    // given array    static void subsetSum(int i, int curr)    {         // Base case        if (i == c.size())        {            ans += curr;            return;        }         // Recursively calling subsetSum        subsetSum(i + 1, curr + c.elementAt(i));        subsetSum(i + 1, curr);    }     // Function to generate the subsets    static void subsetGen(int[] arr, int i, int n)    {         // Base-case        if (i == n)        {             // Finding the sum of all the subsets            // of the generated subset            subsetSum(0, 0);            return;        }         // Recursively accepting and rejecting        // the current number        subsetGen(arr, i + 1, n);        c.add(arr[i]);        subsetGen(arr, i + 1, n);        c.remove(c.size() - 1);    }     // Driver Code    public static void main(String[] args)    {        int[] arr = { 1, 1 };        int n = arr.length;         subsetGen(arr, 0, n);        System.out.println(ans);    }} // This code is contributed by// sanjeev2552

## Python3

 # Python3 implementation of the approach # Function to sum of all subsets# of a given arrayc = []ans = 0 def subsetSum(i, curr):    global ans, c         # Base case    if (i == len(c)):        ans += curr        return     # Recursively calling subsetSum    subsetSum(i + 1, curr + c[i])    subsetSum(i + 1, curr) # Function to generate the subsetsdef subsetGen(arr, i, n):    global ans, c         # Base-case    if (i == n):         # Finding the sum of all the subsets        # of the generated subset        subsetSum(0, 0)        return     # Recursively accepting and rejecting    # the current number    subsetGen(arr, i + 1, n)    c.append(arr[i])    subsetGen(arr, i + 1, n)    del c[-1] # Driver codearr = [1, 1]n = len(arr) subsetGen(arr, 0, n) print(ans) # This code is contributed by Mohit Kumar

## C#

 // C# implementation of the approachusing System;using System.Collections.Generic; class GFG{    static List c = new List();     // To store the readonly ans    static int ans = 0;     // Function to sum of all subsets of a    // given array    static void subsetSum(int i, int curr)    {         // Base case        if (i == c.Count)        {            ans += curr;            return;        }         // Recursively calling subsetSum        subsetSum(i + 1, curr + c[i]);        subsetSum(i + 1, curr);    }     // Function to generate the subsets    static void subsetGen(int[] arr, int i, int n)    {         // Base-case        if (i == n)        {             // Finding the sum of all the subsets            // of the generated subset            subsetSum(0, 0);            return;        }         // Recursively accepting and rejecting        // the current number        subsetGen(arr, i + 1, n);        c.Add(arr[i]);        subsetGen(arr, i + 1, n);        c.RemoveAt(c.Count - 1);    }     // Driver Code    public static void Main(String[] args)    {        int[] arr = { 1, 1 };        int n = arr.Length;         subsetGen(arr, 0, n);        Console.WriteLine(ans);    }} // This code is contributed by Rajput-Ji

## Javascript



Output:

6

Time Complexity: O(n * 2n ), to generate all the subsets where n is the size of the given array
Auxiliary Space: O(n), to store the final answer

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