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++ 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 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 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 |
<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> |
112
Time Complexity: O(n)
Auxiliary Space: O(1)