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Find sum of sum of all sub-sequences

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  • Difficulty Level : Easy
  • Last Updated : 07 Jul, 2022
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Given an array of n integers. The task is to find the sum of each sub-sequence of the array.

Examples : 

Input : arr[] = { 6, 8, 5 }
Output : 76
All subsequence sum are:
{ 6 }, sum = 6
{ 8 }, sum = 8
{ 5 }, sum = 5
{ 6, 8 }, sum = 14
{ 6, 5 }, sum = 11
{ 8, 5 }, sum = 13
{ 6, 8, 5 }, sum = 19
Total sum = 76.

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

Method 1 (brute force): 
Generate all the sub-sequence and find the sum of each sub-sequence.

Method 2 (efficient approach): 
For an array of size n, we have 2^n sub-sequences (including empty) in total. Observe, in total 2n sub-sequences, each element occurs 2n-1 times. 
For example, arr[] = { 5, 6, 7 } 

So, the sum of all sub-sequence will be (sum of all the elements) * 2n-1.

Below is the implementation of this approach: 
 

C++




// C++ program to find sum of all sub-sequences
// of an array.
#include<bits/stdc++.h>
using namespace std;
 
// Return sum of sum of all sub-sequence.
int sum(int arr[], int n)
{
  int ans = 0;
 
  // Finding sum of the array.
  for (int i = 0; i < n; i++)
    ans += arr[i];
 
  return ans * pow(2, n - 1);
}
 
// Driver Code
int main()
{
  int arr[] = { 6, 7, 8 };
  int n = sizeof(arr)/sizeof(arr[0]);
 
  cout << sum(arr, n) << endl;
 
  return 0;
}

Java




// Java program to find sum of
// all sub-sequences of an array.
import java.io.*;
import java.math.*;
 
class GFG {
     
    // Return sum of sum of all sub-sequence.
    static int sum(int arr[], int n)
    {
    int ans = 0;
     
    // Finding sum of the array.
    for (int i = 0; i < n; i++)
        ans += arr[i];
     
    return ans * (int)(Math.pow(2, n - 1));
    }
     
    // Driver Code
    public static void main(String args[])
    {
    int arr[]= { 6, 7, 8 };
    int n = arr.length;
     
    System.out.println(sum(arr, n));
    }
}
     
// This code is contributed by Nikita Tiwari.

Python3




# Python 3 program to find sum of
# all sub-sequences of an array.
 
 
# Return sum of sum of all sub-sequence.
def sm(arr , n) :
    ans = 0
 
    # Finding sum of the array.
    for i in range(0, n) :
        ans = ans + arr[i]
     
    return ans * pow(2, n - 1)
     
     
# Driver Code
arr = [ 6, 7, 8 ]
n=len(arr)
 
print(sm(arr, n))
 
 
# This code is contributed by Nikita Tiwari.

C#




// C# program to find sum of
// all sub-sequences of an array.
using System;
 
class GFG
{
     
    // Return sum of sum of all sub-sequence.
    static int sum(int []arr, int n)
    {
    int ans = 0;
     
    // Finding sum of the array.
    for (int i = 0; i < n; i++)
        ans += arr[i];
     
    return ans * (int)(Math.Pow(2, n - 1));
    }
     
    // Driver Code
    public static void Main()
    {
    int []arr= { 6, 7, 8 };
    int n = arr.Length;
     
    Console.Write(sum(arr, n));
    }
}
     
// This code is contributed by nitin mittal

PHP




<?php
// PHP program to find sum of
// all sub-sequences of an array.
 
// Return sum of sum of
// all sub-sequence.
function sum($arr, $n)
{
    $ans = 0;
 
    // Finding sum of the array.
    for ($i = 0; $i < $n; $i++)
        $ans += $arr[$i];
     
    return $ans * pow(2, $n - 1);
}
 
// Driver Code
$arr = array(6, 7, 8);
$n = sizeof($arr);
echo sum($arr, $n) ;
 
// This code is contributed by nitin mittal.
?>

Javascript




<script>
 
// JavaScript program to find sum of all sub-sequences
// of an array.
 
// Return sum of sum of all sub-sequence.
function sum(arr, n)
{
  var ans = 0;
 
  // Finding sum of the array.
  for (var i = 0; i < n; i++)
    ans += arr[i];
 
  return ans * Math.pow(2, n - 1);
}
 
// Driver Code
var arr = [6, 7, 8];
var n = arr.length;
document.write( sum(arr, n));
 
</script>

Output

84

Time complexity:  O(n)
Auxiliary space: O(1)  

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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