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Sum of all subsequences of an array

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Given an array of n integers. Find the sum of all possible subsequences of an array. 

Examples : 

Input : arr[] = { 1, 2 }
Output : 6
All possible subsequences are {}, {1}, {2} 
and { 1, 2 }

Input : arr[] = { 1, 2, 3 }
Output : 24

We have already discussed two different solutions in below post. 
Sum of all Subarrays | Set 1

In this post a different solution is discussed. Let us take a closer look at the problem and try to find a pattern  

Let a[] = { 1, 2, 3 }

All subsequences are {}, {1}, {2}, {3}, {1, 2}, 
                     {1, 3}, {2, 3}, {1, 2, 3}

So sum of subsequences are 0 + 1 + 2 + 3 + 3 + 
                             4 + 5 + 6 = 24

Here we can observe that in sum every elements 
occurs 4 times. Or in general every element 
will occur 2^(n-1) times. And we can also 
observe that sum of array elements is 6. So
final result will be 6*4.

In general we can find sum of all subsequences by adding all elements of array multiplied by 2(n-1) where n is number of elements in array.

Implementation:

C++




// CPP program to find sum of
// all subarrays of array
#include <bits/stdc++.h>
using namespace std;
  
// To find sum of all subsequences
int findSum(int arr[], int n)
{
    // Sum all array elements
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
  
    // Result is sum * 2^(n-1)
    return sum * (1 << (n - 1));
}
  
// Driver program to test findSum()
int main()
{
    int arr[] = { 1, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findSum(arr, n);
    return 0;
}


Java




// Java program to find sum of
// all subarrays of array
  
public class Main {
    // To find sum of all subsequences
    static int findSum(int arr[], int n)
    {
        // Sum all array elements
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];
  
        // Result is sum * 2^(n-1)
        return sum * (1 << (n - 1));
    }
  
    // Driver program to test findSum()
    public static void main(String[] args)
    {
        int arr[] = { 1, 2 };
        int n = arr.length;
        System.out.print(findSum(arr, n));
    }
}


Python




# Python program to find sum of
# all subarrays of array
  
# To find sum of all subsequences
def findSum(arr, n):
      
    # Sum all array elements
    sum = 0
    for i in range(n):
        sum += arr[i]
   
    # Result is sum * 2^(n-1)
    return sum * (1 << (n - 1))
   
# Driver program to test findSum()
arr = [1, 2]
n = len(arr)
print findSum(arr, n)
  
# This code is submitted by Sachin Bisht


C#




// C# program to find sum of 
// all subarrays of array 
using System;
  
class GFG
{
  
// To find sum of all subsequences
static int findSum(int []arr, int n) 
    // Sum all array elements 
    int sum = 0; 
    for (int i = 0; i < n; i++) 
        sum += arr[i]; 
  
    // Result is sum * 2^(n-1) 
    return sum * (1 << (n - 1)); 
  
// Driver Code
static public void Main ()
{
    int []arr = { 1, 2 }; 
    int n = arr.Length; 
    Console.WriteLine(findSum(arr, n)); 
  
// This code is contributed by ajit


PHP




<?php
// PHP program to find sum of
// all subarrays of array
  
// To find sum of all subsequences
  
function findSum($arr, $n)
{
    // Sum all array elements
    $sum = 0;
    for ($i = 0; $i < $n; $i++)
        $sum += $arr[$i];
  
    // Result is sum * 2^(n-1)
    return $sum * (1 << ($n - 1));
}
  
// Driver Code
$arr = array( 1, 2 );
$n = sizeof($arr);
  
echo findSum($arr, $n);
  
// This code is contributed by ajit
?>


Javascript




<script>
  
// Javascript program to find sum of 
// all subarrays of array 
  
// To find sum of all subsequences
function findSum(arr, n) 
      
    // Sum all array elements 
    let sum = 0; 
    for(let i = 0; i < n; i++) 
        sum += arr[i]; 
  
    // Result is sum * 2^(n-1) 
    return sum * (1 << (n - 1)); 
  
// Driver code
let arr = [ 1, 2 ]; 
let n = arr.length; 
  
document.write(findSum(arr, n)); 
  
// This code is contributed by rameshtravel07
  
</script>


Output

6

 



Last Updated : 18 Sep, 2023
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