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Sum of the sums of all possible subsets

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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 <bits/stdc++.h>
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 <bits/stdc++.h>
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




<script>
// javascript program to find the sum of
// the addition of all possible subsets.   
// Function to find the sum
    // of sum of all the subset
    function sumOfSubset(a , n) {
        var times = parseInt( Math.pow(2, n - 1));
 
        var sum = 0;
 
        for (i = 0; i < n; i++) {
            sum = sum + (a[i] * times);
        }
 
        return sum;
    }
 
    // Driver Code
     
        var a = [ 3, 7 ];
        var n = a.length;
        document.write(sumOfSubset(a, n));
 
// This code is contributed by todaysgaurav
</script>


Output: 

20

 

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



Last Updated : 28 Dec, 2022
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