Skip to content
Related Articles

Related Articles

Sum of subsets of all the subsets of an array | O(2^N)
  • Difficulty Level : Hard
  • Last Updated : 18 Nov, 2019
GeeksforGeeks - Summer Carnival Banner

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 * 2N) 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.



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++




// 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




// 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




# 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#




// 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
Output:
6

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.

My Personal Notes arrow_drop_up
Recommended Articles
Page :