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

Last Updated : 13 Mar, 2022

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++

// 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;
}

Java

// 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

Python3

# 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

// 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

JavaScript

<script>
// javascript implementation of the approach 

// Function to return the required sum 
function sum(arr, n) 
{ 

// Find the sum of the array elements 
var sum = 0; 
for (i = 0; i < n; i++)
{ 
sum += arr[i]; 
} 

// Every element appears 2^(n-1) times 
sum = sum * parseInt(Math.pow(2, n - 1)); 
return sum; 
} 

// Driver code 

var arr = [ 2, 1, 5, 6 ];
var n = arr.length; 
document.write(sum(arr, n)); 

// This code is contributed by Amit Katiyar 

</script>

Output:

Time Complexity: O(n)

Auxiliary Space: O(1)

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

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

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