Sum of subsets of all the subsets of an array | O(2^N)
Last Updated :
27 Jan, 2023
Given an array arr[] of length N, the task is to find the overall sum of subsets of all the subsets of the array.
Examples:
Input: arr[] = {1, 1}
Output: 6
All possible subsets:
a) {} : 0
All the possible subsets of this subset
will be {}, Sum = 0
b) {1} : 1
All the possible subsets of this subset
will be {} and {1}, Sum = 0 + 1 = 1
c) {1} : 1
All the possible subsets of this subset
will be {} and {1}, Sum = 0 + 1 = 1
d) {1, 1} : 4
All the possible subsets of this subset
will be {}, {1}, {1} and {1, 1}, Sum = 0 + 1 + 1 + 2 = 4
Thus, ans = 0 + 1 + 1 + 4 = 6
Input: arr[] = {1, 4, 2, 12}
Output: 513
Approach: In this article, an approach with O(N * 2N) time complexity to solve the given problem will be discussed.
First, generate all the possible subsets of the array. There will be 2N subsets in total. Then for each subset, find the sum of all of its subsets.
For, that it can be observed that in an array of length L, every element will come exactly 2(L - 1) times in the sum of subsets. So, the contribution of each element will be 2(L - 1) times its values.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to sum of all subsets of a
// given array
void subsetSum(vector<int>& c, int& ans)
{
int L = c.size();
int mul = (int)pow(2, L - 1);
for (int i = 0; i < c.size(); i++)
ans += c[i] * mul;
}
// Function to generate the subsets
void subsetGen(int* arr, int i, int n,
int& ans, vector<int>& c)
{
// Base-case
if (i == n) {
// Finding the sum of all the subsets
// of the generated subset
subsetSum(c, ans);
return;
}
// Recursively accepting and rejecting
// the current number
subsetGen(arr, i + 1, n, ans, c);
c.push_back(arr[i]);
subsetGen(arr, i + 1, n, ans, c);
c.pop_back();
}
// Driver code
int main()
{
int arr[] = { 1, 1 };
int n = sizeof(arr) / sizeof(int);
// To store the final ans
int ans = 0;
vector<int> c;
subsetGen(arr, 0, n, ans, c);
cout << ans;
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// To store the final ans
static int ans;
// Function to sum of all subsets of a
// given array
static void subsetSum(Vector<Integer> c)
{
int L = c.size();
int mul = (int)Math.pow(2, L - 1);
for (int i = 0; i < c.size(); i++)
ans += c.get(i) * mul;
}
// Function to generate the subsets
static void subsetGen(int []arr, int i,
int n, Vector<Integer> c)
{
// Base-case
if (i == n)
{
// Finding the sum of all the subsets
// of the generated subset
subsetSum(c);
return;
}
// Recursively accepting and rejecting
// the current number
subsetGen(arr, i + 1, n, c);
c.add(arr[i]);
subsetGen(arr, i + 1, n, c);
c.remove(0);
}
// Driver code
public static void main(String []args)
{
int arr[] = { 1, 1 };
int n = arr.length;
Vector<Integer> c = new Vector<Integer>();
subsetGen(arr, 0, n, c);
System.out.println(ans);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the approach
# store the answer
c = []
ans = 0
# Function to sum of all subsets of a
# given array
def subsetSum():
global ans
L = len(c)
mul = pow(2, L - 1)
i = 0
while ( i < len(c)):
ans += c[i] * mul
i += 1
# Function to generate the subsets
def subsetGen(arr, i, n):
# Base-case
if (i == n) :
# Finding the sum of all the subsets
# of the generated subset
subsetSum()
return
# Recursively accepting and rejecting
# the current number
subsetGen(arr, i + 1, n)
c.append(arr[i])
subsetGen(arr, i + 1, n)
c.pop()
# Driver code
if __name__ == "__main__" :
arr = [ 1, 1 ]
n = len(arr)
subsetGen(arr, 0, n)
print (ans)
# This code is contributed by Arnab Kundu
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// To store the final ans
static int ans;
// Function to sum of all subsets of a
// given array
static void subsetSum(List<int> c)
{
int L = c.Count;
int mul = (int)Math.Pow(2, L - 1);
for (int i = 0; i < c.Count; i++)
ans += c[i] * mul;
}
// Function to generate the subsets
static void subsetGen(int []arr, int i,
int n, List<int> c)
{
// Base-case
if (i == n)
{
// Finding the sum of all the subsets
// of the generated subset
subsetSum(c);
return;
}
// Recursively accepting and rejecting
// the current number
subsetGen(arr, i + 1, n, c);
c.Add(arr[i]);
subsetGen(arr, i + 1, n, c);
c.RemoveAt(0);
}
// Driver code
public static void Main(String []args)
{
int []arr = { 1, 1 };
int n = arr.Length;
List<int> c = new List<int>();
subsetGen(arr, 0, n, c);
Console.WriteLine(ans);
}
}
// This code is contributed by Rajput-Ji
JavaScript
<script>
// javascript implementation of the approach
// To store the final ans
var ans = 0;
// Function to sum of all subsets of a
// given array
function subsetSum( c) {
var L = c.length;
var mul = parseInt( Math.pow(2, L - 1));
for (i = 0; i < c.length; i++)
ans += c[i] * mul;
}
// Function to generate the subsets
function subsetGen(arr , i , n, c) {
// Base-case
if (i == n) {
// Finding the sum of all the subsets
// of the generated subset
subsetSum(c);
return;
}
// Recursively accepting and rejecting
// the current number
subsetGen(arr, i + 1, n, c);
c.push(arr[i]);
subsetGen(arr, i + 1, n, c);
c.pop(0);
}
// Driver code
var arr = [ 1, 1 ];
var n = arr.length;
var c = [];
subsetGen(arr, 0, n, c);
document.write(ans);
// This code is contributed by todaysgaurav
</script>
Time Complexity: O(2^n), where n is the size of the given array.
Subset generation takes O(2^n) time as there are 2^n subsets of a given set.
Space Complexity: O(n).
Recursion stack will be used which will take O(n) space.
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