Given an array arr[] of N integers, the task is to find the sum of values of all possible non-empty subsets of array the given array.
Examples:
Input: arr[] = {2, 3}
Output: 10
All non-empty subsets are {2}, {3} and {2, 3}
Total sum = 2 + 3 + 2 + 3 = 10
Input: arr[] = {2, 1, 5, 6}
Output: 112
Approach: It can be observed that when all the elements are added from all the possible subsets then each element of the original array appears in 2(N – 1) times. Which means contribution of any element arr[i] in the final answer will be arr[i] * 2(N – 1). So, the required answer will be (arr[0] + arr[1] + arr[2] + … + arr[N – 1]) * 2(N – 1).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the required sumint sum(int arr[], int n){ // Find the sum of the array elements int sum = 0; for (int i = 0; i < n; i++) { sum += arr[i]; } // Every element appears 2^(n-1) times sum = sum * pow(2, n - 1); return sum;}// Driver codeint main(){ int arr[] = { 2, 1, 5, 6 }; int n = sizeof(arr) / sizeof(int); cout << sum(arr, n); return 0;} |
Java
// Java implementation of the approach class GFG{ // Function to return the required sum static int sum(int arr[], int n) { // Find the sum of the array elements int sum = 0; for (int i = 0; i < n; i++) { sum += arr[i]; } // Every element appears 2^(n-1) times sum = sum * (int)Math.pow(2, n - 1); return sum; } // Driver code public static void main (String[] args) { int arr[] = { 2, 1, 5, 6 }; int n = arr.length; System.out.println(sum(arr, n)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach# Function to return the required sumdef sum( arr, n): # Find the sum of the array elements sum = 0 for i in arr : sum += i # Every element appears 2^(n-1) times sum = sum * pow(2, n - 1) return sum# Driver codearr = [ 2, 1, 5, 6 ]n = len(arr)print(sum(arr, n))# This code is contributed by Arnab Kundu |
C#
// C# implementation of the approach using System;class GFG{ // Function to return the required sum static int sum(int[] arr, int n) { // Find the sum of the array elements int sum = 0; for (int i = 0; i < n; i++) { sum += arr[i]; } // Every element appears 2^(n-1) times sum = sum * (int)Math.Pow(2, n - 1); return sum; } // Driver code public static void Main () { int[] arr = { 2, 1, 5, 6 }; int n = arr.Length; Console.WriteLine(sum(arr, n)); } }// This code is contributed by CodeMech |
Javascript
<script>// javascript implementation of the approach // Function to return the required sum function sum(arr, n) { // Find the sum of the array elements var sum = 0; for (i = 0; i < n; i++){ sum += arr[i]; } // Every element appears 2^(n-1) times sum = sum * parseInt(Math.pow(2, n - 1)); return sum; } // Driver code var arr = [ 2, 1, 5, 6 ];var n = arr.length; document.write(sum(arr, n)); // This code is contributed by Amit Katiyar </script> |
Output:
112
Time Complexity: O(n)
Auxiliary Space: O(1)
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