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++
// 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 i,
int& ans, int curr)
{
// Base case
if (i == c.size()) {
ans += curr;
return;
}
// Recursively calling subsetSum
subsetSum(c, i + 1, ans, curr + c[i]);
subsetSum(c, i + 1, ans, curr);
}
// 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, 0, ans, 0);
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 implementation of the approach
import java.util.*;
class GFG
{
static Vector<Integer> c = new Vector<>();
// To store the final ans
static int ans = 0;
// Function to sum of all subsets of a
// given array
static void subsetSum(int i, int curr)
{
// Base case
if (i == c.size())
{
ans += curr;
return;
}
// Recursively calling subsetSum
subsetSum(i + 1, curr + c.elementAt(i));
subsetSum(i + 1, curr);
}
// Function to generate the subsets
static void subsetGen(int[] arr, int i, int n)
{
// Base-case
if (i == n)
{
// Finding the sum of all the subsets
// of the generated subset
subsetSum(0, 0);
return;
}
// Recursively accepting and rejecting
// the current number
subsetGen(arr, i + 1, n);
c.add(arr[i]);
subsetGen(arr, i + 1, n);
c.remove(c.size() - 1);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 1 };
int n = arr.length;
subsetGen(arr, 0, n);
System.out.println(ans);
}
}
// This code is contributed by
// sanjeev2552
# Python3 implementation of the approach
# Function to sum of all subsets
# of a given array
c = []
ans = 0
def subsetSum(i, curr):
global ans, c
# Base case
if (i == len(c)):
ans += curr
return
# Recursively calling subsetSum
subsetSum(i + 1, curr + c[i])
subsetSum(i + 1, curr)
# Function to generate the subsets
def subsetGen(arr, i, n):
global ans, c
# Base-case
if (i == n):
# Finding the sum of all the subsets
# of the generated subset
subsetSum(0, 0)
return
# Recursively accepting and rejecting
# the current number
subsetGen(arr, i + 1, n)
c.append(arr[i])
subsetGen(arr, i + 1, n)
del c[-1]
# Driver code
arr = [1, 1]
n = len(arr)
subsetGen(arr, 0, n)
print(ans)
# This code is contributed by Mohit Kumar
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static List<int> c = new List<int>();
// To store the readonly ans
static int ans = 0;
// Function to sum of all subsets of a
// given array
static void subsetSum(int i, int curr)
{
// Base case
if (i == c.Count)
{
ans += curr;
return;
}
// Recursively calling subsetSum
subsetSum(i + 1, curr + c[i]);
subsetSum(i + 1, curr);
}
// Function to generate the subsets
static void subsetGen(int[] arr, int i, int n)
{
// Base-case
if (i == n)
{
// Finding the sum of all the subsets
// of the generated subset
subsetSum(0, 0);
return;
}
// Recursively accepting and rejecting
// the current number
subsetGen(arr, i + 1, n);
c.Add(arr[i]);
subsetGen(arr, i + 1, n);
c.RemoveAt(c.Count - 1);
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 1, 1 };
int n = arr.Length;
subsetGen(arr, 0, n);
Console.WriteLine(ans);
}
}
// This code is contributed by Rajput-Ji
<script>
// Javascript implementation of the approach
var ans = 0;
var c = [];
// Function to sum of all subsets of a
// given array
function subsetSum(i, curr)
{
// Base case
if (i == c.length) {
ans += curr;
return;
}
// Recursively calling subsetSum
subsetSum( i + 1, curr + c[i]);
subsetSum(i + 1, curr);
}
// Function to generate the subsets
function subsetGen(arr, i, n, ans)
{
// Base-case
if (i == n) {
// Finding the sum of all the subsets
// of the generated subset
subsetSum(0, ans, 0);
return;
}
// Recursively accepting and rejecting
// the current number
subsetGen(arr, i + 1, n, ans);
c.push(arr[i]);
subsetGen(arr, i + 1, n, ans);
c.pop();
}
// Driver code
var arr = [1, 1 ];
var n = arr.length;
// To store the final ans
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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