Sum of subsets of all the subsets of an array | O(3^N)
Last Updated :
10 Jun, 2022
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(3N) 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 subset.
Now, let’s understand the time-complexity of this solution.
There are NCk subsets of length K and time to find the subsets of an array of length K is 2K.
Total time = (NC1 * 21) + (NC2 * 22) + … + (NCk * K) = 3K
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void subsetSum(vector< int >& c, int i,
int & ans, int curr)
{
if (i == c.size()) {
ans += curr;
return ;
}
subsetSum(c, i + 1, ans, curr + c[i]);
subsetSum(c, i + 1, ans, curr);
}
void subsetGen( int * arr, int i, int n,
int & ans, vector< int >& c)
{
if (i == n) {
subsetSum(c, 0, ans, 0);
return ;
}
subsetGen(arr, i + 1, n, ans, c);
c.push_back(arr[i]);
subsetGen(arr, i + 1, n, ans, c);
c.pop_back();
}
int main()
{
int arr[] = { 1, 1 };
int n = sizeof (arr) / sizeof ( int );
int ans = 0;
vector< int > c;
subsetGen(arr, 0, n, ans, c);
cout << ans;
return 0;
}
|
Java
import java.util.*;
class GFG
{
static Vector<Integer> c = new Vector<>();
static int ans = 0 ;
static void subsetSum( int i, int curr)
{
if (i == c.size())
{
ans += curr;
return ;
}
subsetSum(i + 1 , curr + c.elementAt(i));
subsetSum(i + 1 , curr);
}
static void subsetGen( int [] arr, int i, int n)
{
if (i == n)
{
subsetSum( 0 , 0 );
return ;
}
subsetGen(arr, i + 1 , n);
c.add(arr[i]);
subsetGen(arr, i + 1 , n);
c.remove(c.size() - 1 );
}
public static void main(String[] args)
{
int [] arr = { 1 , 1 };
int n = arr.length;
subsetGen(arr, 0 , n);
System.out.println(ans);
}
}
|
Python3
c = []
ans = 0
def subsetSum(i, curr):
global ans, c
if (i = = len (c)):
ans + = curr
return
subsetSum(i + 1 , curr + c[i])
subsetSum(i + 1 , curr)
def subsetGen(arr, i, n):
global ans, c
if (i = = n):
subsetSum( 0 , 0 )
return
subsetGen(arr, i + 1 , n)
c.append(arr[i])
subsetGen(arr, i + 1 , n)
del c[ - 1 ]
arr = [ 1 , 1 ]
n = len (arr)
subsetGen(arr, 0 , n)
print (ans)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static List< int > c = new List< int >();
static int ans = 0;
static void subsetSum( int i, int curr)
{
if (i == c.Count)
{
ans += curr;
return ;
}
subsetSum(i + 1, curr + c[i]);
subsetSum(i + 1, curr);
}
static void subsetGen( int [] arr, int i, int n)
{
if (i == n)
{
subsetSum(0, 0);
return ;
}
subsetGen(arr, i + 1, n);
c.Add(arr[i]);
subsetGen(arr, i + 1, n);
c.RemoveAt(c.Count - 1);
}
public static void Main(String[] args)
{
int [] arr = { 1, 1 };
int n = arr.Length;
subsetGen(arr, 0, n);
Console.WriteLine(ans);
}
}
|
Javascript
<script>
var ans = 0;
var c = [];
function subsetSum(i, curr)
{
if (i == c.length) {
ans += curr;
return ;
}
subsetSum( i + 1, curr + c[i]);
subsetSum(i + 1, curr);
}
function subsetGen(arr, i, n, ans)
{
if (i == n) {
subsetSum(0, ans, 0);
return ;
}
subsetGen(arr, i + 1, n, ans);
c.push(arr[i]);
subsetGen(arr, i + 1, n, ans);
c.pop();
}
var arr = [1, 1 ];
var n = arr.length;
var ans = 0;
var c = [];
subsetGen(arr, 0, n, ans);
document.write( ans);
</script>
|
Time Complexity: O(n * 2n ), to generate all the subsets where n is the size of the given array
Auxiliary Space: O(n), to store the final answer
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