Given an array of size N. The task is to find the product of values of all possible non-empty subsets of the given array.
Examples:
Input: N = 2, arr[] = {3, 7}
Output: 441
All non empty subsets are:
3
7
3, 7
Product = 3 * 7 * 3 * 7 = 441
Input: N = 1, arr[] = {4}
Output: 4
Approach: On careful observation it can be deduced that number of occurrences of every element across all subsets is 2N-1. Therefore, in the final product, every element of the array will be multiplied to itself by 2N-1 times.
Product = a[0]2N-1*a[1]2N-1********a[N-1]2N-1
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to find product of all elements// in all subsetsint product(int a[], int n){ int ans = 1; int val = pow(2, n - 1); for (int i = 0; i < n; i++) { ans *= pow(a[i], val); } return ans;}// Driver Codeint main(){ int n = 2; int a[] = { 3, 7 }; cout << product(a, n); return 0;} |
Java
// Java implementation of the approach class GFG { // Function to find product of all elements // in all subsets static int product(int a[], int n) { int ans = 1; int val = (int)Math.pow(2, n - 1); for (int i = 0; i < n; i++) { ans *= (int)Math.pow(a[i], val); } return ans; } // Driver Code public static void main (String[] args) { int n = 2; int a[] = { 3, 7 }; System.out.println(product(a, n)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach# Function to find product of # all elements in all subsetsdef product(a, n): ans = 1 val = pow(2, n - 1) for i in range(n): ans *= pow(a[i], val) return ans# Driver Coden = 2a = [3, 7]print(product(a, n))# This code is contributed by mohit kumar |
C#
// C# implementation of the approach using System;class GFG { // Function to find product of all elements // in all subsets static int product(int []a, int n) { int ans = 1; int val = (int)Math.Pow(2, n - 1); for (int i = 0; i < n; i++) { ans *= (int)Math.Pow(a[i], val); } return ans; } // Driver Code public static void Main () { int n = 2; int []a = { 3, 7 }; Console.WriteLine(product(a, n)); } }// This code is contributed by anuj_67.. |
Javascript
<script>// Javascript implementation of the approach// Function to find product of all elements// in all subsetsfunction product(a, n){ var ans = 1; var val = Math.pow(2, n - 1); for (var i = 0; i < n; i++) { ans *= Math.pow(a[i], val); } return ans;}// Driver Codevar n = 2;a = [ 3, 7 ] document.write(product(a, n));</script> |
Output:
441
Time Complexity: O(N)
Auxiliary Space: O(1)
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