Sum of the sums of all possible subsets
Last Updated :
28 Dec, 2022
Given an array a of size N. The task is to find the sum of the sums of all possible subsets.
Examples:
Input: a[] = {3, 7}
Output: 20
The subsets are: {3} {7} {3, 7}
{3, 7} = 10
{3} = 3
{7} = 7
10 + 3 + 7 = 20
Input: a[] = {10, 16, 14, 9}
Output: 392
Naive Approach: A naive approach is to find all the subsets using power set and then summate all the possible subsets to get the answer.
C++
// C++ program to check if there is a subset
// with sum divisible by m.
#include <bits/stdc++.h>
using namespace std;
int helper(int N, int nums[], int sum, int idx)
{
// if we reach last index
if (idx == N) {
// and if the sum mod m is zero
return sum;
}
// 2 choices - to pick or to not pick
int picked = helper(N, nums, sum + nums[idx], idx + 1);
int notPicked = helper(N, nums, sum, idx + 1);
return picked + notPicked;
}
int sumOfSubset(int arr[], int n)
{
return helper(n, arr, 0, 0);
}
// Driver code
int main()
{
int arr[] = { 3, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << sumOfSubset(arr, n);
return 0;
}
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
static int helper(int N, int nums[], int sum, int idx)
{
// if we reach last index
if (idx == N) {
// and if the sum mod m is zero
return sum;
}
// 2 choices - to pick or to not pick
int picked
= helper(N, nums, sum + nums[idx], idx + 1);
int notPicked = helper(N, nums, sum, idx + 1);
return picked + notPicked;
}
static int sumOfSubset(int arr[], int n)
{
return helper(n, arr, 0, 0);
}
public static void main(String[] args)
{
int arr[] = { 3, 7 };
int n = arr.length;
System.out.println(sumOfSubset(arr, n));
}
}
// This code is contributed by aadityaburujwale.
# Python program to check if there is a subset
# with sum divisible by m.
def helper(N, nums, sum, idx):
# if we reach last index
if idx == N:
# and if the sum mod m is zero
return sum
# 2 choices - to pick or to not pick
picked = helper(N, nums, sum + nums[idx], idx + 1)
not_picked = helper(N, nums, sum, idx + 1)
return picked + not_picked
def sum_of_subset(arr, n):
return helper(n, arr, 0, 0)
# Test the program
arr = [3, 7]
n = len(arr)
print(sum_of_subset(arr, n))
# This code is contributed by divyansh2212
/*package whatever //do not write package name here */
using System;
class GFG {
static int helper(int N, int[] nums, int sum, int idx)
{
// if we reach last index
if (idx == N) {
// and if the sum mod m is zero
return sum;
}
// 2 choices - to pick or to not pick
int picked
= helper(N, nums, sum + nums[idx], idx + 1);
int notPicked = helper(N, nums, sum, idx + 1);
return picked + notPicked;
}
static int sumOfSubset(int[] arr, int n)
{
return helper(n, arr, 0, 0);
}
public static void Main()
{
int[] arr = { 3, 7 };
int n = arr.Length;
Console.WriteLine(sumOfSubset(arr, n));
}
}
// This code is contributed by Samim Hossain Mondal.
// JS program to check if there is a subset
// with sum divisible by m.
function helper(N, nums, sum, idx)
{
// if we reach last index
if (idx == N)
{
// and if the sum mod m is zero
return sum;
}
// 2 choices - to pick or to not pick
let picked = helper(N, nums, sum + nums[idx], idx + 1);
let notPicked = helper(N, nums, sum, idx + 1);
return picked + notPicked;
}
function sumOfSubset(arr, n)
{
return helper(n, arr, 0, 0);
}
// Driver code
let arr = [ 3, 7 ];
let n = arr.length;
console.log(sumOfSubset(arr, n));
// This code is contributed by akashish__
Time Complexity: O(2N)
Space Complexity: O(N) because of Recursion Stack Space
Efficient Approach: An efficient approach is to solve the problem using observation. If we write all the subsequences, a common point of observation is that each number appears 2(N - 1) times in a subset and hence will lead to the 2(N-1) as the contribution to the sum. Iterate through the array and add (arr[i] * 2N-1) to the answer.
Below is the implementation of the above approach:
C++
// C++ program to find the sum of
// the addition of all possible subsets.
#include <bits/stdc++.h>
using namespace std;
// Function to find the sum
// of sum of all the subset
int sumOfSubset(int a[], int n)
{
int times = pow(2, n - 1);
int sum = 0;
for (int i = 0; i < n; i++) {
sum = sum + (a[i] * times);
}
return sum;
}
// Driver Code
int main()
{
int a[] = { 3, 7 };
int n = sizeof(a) / sizeof(a[0]);
cout << sumOfSubset(a, n);
}
// Java program to find the sum of
// the addition of all possible subsets.
class GFG
{
// Function to find the sum
// of sum of all the subset
static int sumOfSubset(int []a, int n)
{
int times = (int)Math.pow(2, n - 1);
int sum = 0;
for (int i = 0; i < n; i++)
{
sum = sum + (a[i] * times);
}
return sum;
}
// Driver Code
public static void main(String[] args)
{
int []a = { 3, 7 };
int n = a.length;
System.out.println(sumOfSubset(a, n));
}
}
// This code is contributed by 29AjayKumar
# Python3 program to find the Sum of
# the addition of all possible subsets.
# Function to find the sum
# of sum of all the subset
def SumOfSubset(a, n):
times = pow(2, n - 1)
Sum = 0
for i in range(n):
Sum = Sum + (a[i] * times)
return Sum
# Driver Code
a = [3, 7]
n = len(a)
print(SumOfSubset(a, n))
# This code is contributed by Mohit Kumar
// C# program to find the sum of
// the addition of all possible subsets.
using System;
class GFG
{
// Function to find the sum
// of sum of all the subset
static int sumOfSubset(int []a, int n)
{
int times = (int)Math.Pow(2, n - 1);
int sum = 0;
for (int i = 0; i < n; i++)
{
sum = sum + (a[i] * times);
}
return sum;
}
// Driver Code
public static void Main()
{
int []a = { 3, 7 };
int n = a.Length;
Console.Write(sumOfSubset(a, n));
}
}
// This code is contributed by Nidhi
<script>
// javascript program to find the sum of
// the addition of all possible subsets.
// Function to find the sum
// of sum of all the subset
function sumOfSubset(a , n) {
var times = parseInt( Math.pow(2, n - 1));
var sum = 0;
for (i = 0; i < n; i++) {
sum = sum + (a[i] * times);
}
return sum;
}
// Driver Code
var a = [ 3, 7 ];
var n = a.length;
document.write(sumOfSubset(a, n));
// This code is contributed by todaysgaurav
</script>
Time Complexity: O(N)
Space Complexity: O(1)
Note: If N is large, the answer can overflow, thereby use larger data-type.
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