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

  • Last Updated : 10 Mar, 2021

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
 

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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#




// 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: 
112

 




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