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 arrayvoid 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 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, 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 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{ 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
# Python3 implementation of the approach# Function to sum of all subsets # of a given arrayc = []ans = 0def 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 subsetsdef 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 codearr = [1, 1]n = len(arr)subsetGen(arr, 0, n)print(ans)# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing 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 |
Javascript
<script>// Javascript implementation of the approachvar ans = 0;var c = [];// Function to sum of all subsets of a// given arrayfunction 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 subsetsfunction 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 codevar arr = [1, 1 ];var n = arr.length;// To store the final ansvar ans = 0;var c = [];subsetGen(arr, 0, n, ans);document.write( ans);</script> |
6
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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