Sum of subsets of all the subsets of an array | O(2^N)

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: 6 
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(N * 2^N) time complexity to solve the given problem will be discussed. 
First, generate all the possible subsets of the array. There will be 2^N subsets in total. Then for each subset, find the sum of all of its subsets.
For, that it can be observed that in an array of length L, every element will come exactly 2^{(L - 1)} times in the sum of subsets. So, the contribution of each element will be 2^{(L - 1)} times its values.
Below is the implementation of the above approach: 
 

// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to sum of all subsets of a // given array void subsetSum(vector<int>& c, int& ans) {  int L = c.size();  int mul = (int)pow(2, L - 1);  for (int i = 0; i < c.size(); i++)  ans += c[i] * mul; } // Function to generate the subsets void subsetGen(int* arr, int i, int n,  int& ans, vector<int>& c) {  // Base-case  if (i == n) {  // Finding the sum of all the subsets  // of the generated subset  subsetSum(c, ans);  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 code int main() {  int arr[] = { 1, 1 };  int n = sizeof(arr) / sizeof(int);  // To store the final ans  int ans = 0;  vector<int> c;  subsetGen(arr, 0, n, ans, c);  cout << ans;  return 0; } 

// Java implementation of the approach import java.util.*; class GFG  { // To store the final ans static int ans; // Function to sum of all subsets of a // given array static void subsetSum(Vector<Integer> c) {  int L = c.size();  int mul = (int)Math.pow(2, L - 1);  for (int i = 0; i < c.size(); i++)  ans += c.get(i) * mul; } // Function to generate the subsets static void subsetGen(int []arr, int i,   int n, Vector<Integer> c) {  // Base-case  if (i == n)   {  // Finding the sum of all the subsets  // of the generated subset  subsetSum(c);  return;  }  // Recursively accepting and rejecting  // the current number  subsetGen(arr, i + 1, n, c);  c.add(arr[i]);  subsetGen(arr, i + 1, n, c);  c.remove(0); } // Driver code public static void main(String []args)  {  int arr[] = { 1, 1 };  int n = arr.length;  Vector<Integer> c = new Vector<Integer>();  subsetGen(arr, 0, n, c);  System.out.println(ans); } } // This code is contributed by 29AjayKumar 

# Python3 implementation of the approach # store the answer c = [] ans = 0 # Function to sum of all subsets of a # given array def subsetSum(): global ans L = len(c) mul = pow(2, L - 1) i = 0 while ( i < len(c)): ans += c[i] * mul i += 1 # Function to generate the subsets def subsetGen(arr, i, n): # Base-case if (i == n) : # Finding the sum of all the subsets # of the generated subset subsetSum() return # Recursively accepting and rejecting # the current number subsetGen(arr, i + 1, n) c.append(arr[i]) subsetGen(arr, i + 1, n) c.pop() # Driver code  if __name__ == "__main__" : arr = [ 1, 1 ] n = len(arr) subsetGen(arr, 0, n) print (ans) # This code is contributed by Arnab Kundu 

// C# implementation of the approach using System; using System.Collections.Generic;  class GFG  { // To store the final ans static int ans; // Function to sum of all subsets of a // given array static void subsetSum(List<int> c) {  int L = c.Count;  int mul = (int)Math.Pow(2, L - 1);  for (int i = 0; i < c.Count; i++)  ans += c[i] * mul; } // Function to generate the subsets static void subsetGen(int []arr, int i,   int n, List<int> c) {  // Base-case  if (i == n)   {  // Finding the sum of all the subsets  // of the generated subset  subsetSum(c);  return;  }  // Recursively accepting and rejecting  // the current number  subsetGen(arr, i + 1, n, c);  c.Add(arr[i]);  subsetGen(arr, i + 1, n, c);  c.RemoveAt(0); } // Driver code public static void Main(String []args)  {  int []arr = { 1, 1 };  int n = arr.Length;  List<int> c = new List<int>();  subsetGen(arr, 0, n, c);  Console.WriteLine(ans); } } // This code is contributed by Rajput-Ji 

<script> // javascript implementation of the approach  // To store the final ans  var ans = 0;  // Function to sum of all subsets of a  // given array  function subsetSum( c) {  var L = c.length;  var mul = parseInt( Math.pow(2, L - 1));  for (i = 0; i < c.length; i++)  ans += c[i] * mul;  }  // Function to generate the subsets  function subsetGen(arr , i , n, c) {  // Base-case  if (i == n) {  // Finding the sum of all the subsets  // of the generated subset  subsetSum(c);  return;  }  // Recursively accepting and rejecting  // the current number  subsetGen(arr, i + 1, n, c);  c.push(arr[i]);  subsetGen(arr, i + 1, n, c);  c.pop(0);  }  // Driver code    var arr = [ 1, 1 ];  var n = arr.length;  var c = [];  subsetGen(arr, 0, n, c);  document.write(ans); // This code is contributed by todaysgaurav  </script> 

6

Time Complexity: O(2^n), where n is the size of the given array.

Subset generation takes O(2^n) time as there are 2^n subsets of a given set.

Space Complexity: O(n).

Recursion stack will be used which will take O(n) space.