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Sum of subsets of all the subsets of an array | O(2^N)

 4 years ago
source link: https://www.geeksforgeeks.org/sum-of-subsets-of-all-the-subsets-of-an-array-o2n/
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Sum of subsets of all the subsets of an array
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Sum of subsets of all the subsets of an array | O(2^N)
  • Difficulty Level : Hard
  • Last Updated : 18 Nov, 2019

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 subset.

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:

  • Python3

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// 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;
}
Output: 
6

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