Sum of values of all possible non-empty subsets of the given array

Given an array arr[] of N integers, the task is to find the sum of values of all possible non-empty subsets of array the given array.

Examples:

Input: arr[] = {2, 3}
Output: 10
All non-empty subsets are {2}, {3} and {2, 3}
Total sum = 2 + 3 + 2 + 3 = 10



Input: arr[] = {2, 1, 5, 6}
Output: 112

Approach: It can be observed that when all the elements are added from all the possible subsets then each element of the original array appears in 2(N – 1) times. Which means contribution of any element arr[i] in the final answer will be arr[i] * 2(N – 1). So, the required answer will be (arr[0] + arr[1] + arr[2] + … + arr[N – 1]) * 2(N – 1).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the required sum
int sum(int arr[], int n)
{
  
    // Find the sum of the array elements
    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += arr[i];
    }
  
    // Every element appears 2^(n-1) times
    sum = sum * pow(2, n - 1);
    return sum;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 1, 5, 6 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << sum(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the approach 
class GFG
{
  
    // Function to return the required sum 
    static int sum(int arr[], int n) 
    
      
        // Find the sum of the array elements 
        int sum = 0
        for (int i = 0; i < n; i++)
        
            sum += arr[i]; 
        
      
        // Every element appears 2^(n-1) times 
        sum = sum * (int)Math.pow(2, n - 1); 
        return sum; 
    
      
    // Driver code 
    public static void main (String[] args) 
    
        int arr[] = { 2, 1, 5, 6 }; 
        int n = arr.length; 
        System.out.println(sum(arr, n)); 
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the approach
  
# Function to return the required sum
def sum( arr, n):
  
    # Find the sum of the array elements
    sum = 0
    for i in arr : 
        sum += i
      
    # Every element appears 2^(n-1) times
    sum = sum * pow(2, n - 1)
    return sum
  
# Driver code
arr = [ 2, 1, 5, 6 ]
n = len(arr)
  
print(sum(arr, n))
  
# This code is contributed by Arnab Kundu 

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C#

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// C# implementation of the approach 
using System;
class GFG
{
  
    // Function to return the required sum 
    static int sum(int[] arr, int n) 
    
      
        // Find the sum of the array elements 
        int sum = 0; 
        for (int i = 0; i < n; i++)
        
            sum += arr[i]; 
        
      
        // Every element appears 2^(n-1) times 
        sum = sum * (int)Math.Pow(2, n - 1); 
        return sum; 
    
      
    // Driver code 
    public static void Main () 
    
        int[] arr = { 2, 1, 5, 6 }; 
        int n = arr.Length; 
        Console.WriteLine(sum(arr, n)); 
    
}
  
// This code is contributed by CodeMech

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Output:

112


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