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 arrayvoid 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 subsetsvoid 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 codeint 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 approachimport java.util.*;class GFG {// To store the final ansstatic int ans;// Function to sum of all subsets of a// given arraystatic 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 subsetsstatic 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 codepublic 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 answerc = []ans = 0# Function to sum of all subsets of a# given arraydef 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 subsetsdef 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 approachusing System;using System.Collections.Generic; class GFG {// To store the final ansstatic int ans;// Function to sum of all subsets of a// given arraystatic 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 subsetsstatic 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 codepublic 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> |
6
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.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
