Given an array a[] of size N. The value of a subset is the product of primes in that subset. A non-prime is considered to be 1 while finding value by-product. The task is to find the product of the value of all possible subsets.
Examples:
Input: a[] = {3, 7}
Output: 20
The subsets are: {3} {7} {3, 7}
{3, 7} = 3 * 7 = 21
{3} = 3
{7} = 7
21 * 3 * 7 = 441
Input: a[] = {10, 2, 14, 3}
Output: 1679616
Naive Approach: A naive approach is to find all the subsets using power set and then find the product by multiplying all the values of the subset. Prime can be checked using Sieve.
Time Complexity: O(2N)
Efficient Approach: An efficient approach is to solve the problem using observation. If we write all the subsequences, a common point of observation is that each number appears 2(N – 1) times in a subset and hence will lead to the 2(N-1) as the contribution to the product. Iterate through the array and check if the element in the array is prime or not. If it prime, then its contribution is arr[i]2(N-1) times to the answer.
Below is the implementation of the above approach:
C++
// C++ program to find the product of// the multiplication of// prime numbers in all possible subsets.#include <bits/stdc++.h>using namespace std;// Sieve method to check prime or notvoid sieve(int n, vector<bool>& prime){ // Initially mark all primes for (int i = 2; i <= n; i++) prime[i] = true; prime[0] = prime[1] = false; // Iterate and mark all the // non primes as false for (int i = 2; i <= n; i++) { if (prime[i]) { // Multiples of prime marked as false for (int j = i * i; j <= n; j += i) { prime[j] = false; } } }}// Function to find the sum// of sum of all the subsetint sumOfSubset(int a[], int n){ // Get the maximum element int maxi = *max_element(a, a + n); // Declare a sieve array vector<bool> prime(maxi + 1); // Sieve function called sieve(maxi, prime); // Number of times an element // contributes to the answer int times = pow(2, n - 1); int sum = 1; // Iterate and check for (int i = 0; i < n; i++) { // If prime if (prime[a[i]]) sum = sum * (pow(a[i], times)); // Contribution } return sum;}// Driver Codeint main(){ int a[] = { 3, 7 }; int n = sizeof(a) / sizeof(a[0]); cout << sumOfSubset(a, n);} |
Java
// Java program to find the product of// the multiplication of// prime numbers in all possible subsets.import java.util.*;class GFG {// Sieve method to check prime or notstatic void sieve(int n, boolean []prime){ // Initially mark all primes for (int i = 2; i <= n; i++) prime[i] = true; prime[0] = prime[1] = false; // Iterate and mark all the // non primes as false for (int i = 2; i <= n; i++) { if (prime[i]) { // Multiples of prime marked as false for (int j = i * i; j <= n; j += i) { prime[j] = false; } } }}// Function to find the sum// of sum of all the subsetstatic int sumOfSubset(int a[], int n){ // Get the maximum element int maxi = Arrays.stream(a).max().getAsInt(); // Declare a sieve array boolean []prime = new boolean[maxi + 1]; // Sieve function called sieve(maxi, prime); // Number of times an element // contributes to the answer int times = (int) Math.pow(2, n - 1); int sum = 1; // Iterate and check for (int i = 0; i < n; i++) { // If prime if (prime[a[i]]) sum = (int) (sum * (Math.pow(a[i], times))); } return sum;}// Driver Codepublic static void main(String[] args) { int a[] = { 3, 7 }; int n = a.length; System.out.println(sumOfSubset(a, n));}}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to find the product of# the multiplication of# prime numbers in all possible subsets.prime = [True for i in range(100)]# Sieve method to check prime or notdef sieve(n, prime): # Initially mark all primes for i in range(1, n + 1): prime[i] = True prime[0] = prime[1] = False # Iterate and mark all the # non primes as false for i in range(2, n + 1): if (prime[i]): # Multiples of prime marked as false for j in range(2 * i, n + 1, i): prime[j] = False# Function to find the Sum# of Sum of all the subsetdef SumOfSubset(a, n): # Get the maximum element maxi = max(a) # Declare a sieve array # Sieve function called sieve(maxi, prime) # Number of times an element # contributes to the answer times = pow(2, n - 1) Sum = 1 # Iterate and check for i in range(n): # If prime if (prime[a[i]]): Sum = Sum * (pow(a[i], times)) # Contribution return Sum# Driver Codea = [3, 7]n = len(a)print(SumOfSubset(a, n))# This code is contributed # by Mohit Kumar |
C#
// C# program to find the product of// the multiplication of// prime numbers in all possible subsets.using System;using System.Linq;class GFG {// Sieve method to check prime or notstatic void sieve(int n, Boolean []prime){ // Initially mark all primes for (int i = 2; i <= n; i++) prime[i] = true; prime[0] = prime[1] = false; // Iterate and mark all the // non primes as false for (int i = 2; i <= n; i++) { if (prime[i]) { // Multiples of prime marked as false for (int j = i * i; j <= n; j += i) { prime[j] = false; } } }}// Function to find the sum// of sum of all the subsetstatic int sumOfSubset(int []a, int n){ // Get the maximum element int maxi = a.Max(); // Declare a sieve array Boolean []prime = new Boolean[maxi + 1]; // Sieve function called sieve(maxi, prime); // Number of times an element // contributes to the answer int times = (int) Math.Pow(2, n - 1); int sum = 1; // Iterate and check for (int i = 0; i < n; i++) { // If prime if (prime[a[i]]) sum = (int) (sum * (Math.Pow(a[i], times))); } return sum;}// Driver Codepublic static void Main(String[] args) { int []a = { 3, 7 }; int n = a.Length; Console.WriteLine(sumOfSubset(a, n));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// javascript program to find the product of// the multiplication of// prime numbers in all possible subsets. // Sieve method to check prime or not function sieve(n, prime) { // Initially mark all primes for (var i = 2; i <= n; i++) prime[i] = true; prime[0] = prime[1] = false; // Iterate and mark all the // non primes as false for (i = 2; i <= n; i++) { if (prime[i]) { // Multiples of prime marked as false for (j = i * i; j <= n; j += i) { prime[j] = false; } } } } // Function to find the sum // of sum of all the subset function sumOfSubset(a , n) { // Get the maximum element var maxi = Math.max.apply(Math,a); // Declare a sieve array var prime =Array(maxi + 1).fill(0); // Sieve function called sieve(maxi, prime); // Number of times an element // contributes to the answer var times = parseInt( Math.pow(2, n - 1)); var sum = 1; // Iterate and check for (i = 0; i < n; i++) { // If prime if (prime[a[i]]) sum = parseInt( (sum * (Math.pow(a[i], times)))); } return sum; } // Driver Code var a = [ 3, 7 ]; var n = a.length; document.write(sumOfSubset(a, n));// This code is contributed by umadevi9616 </script> |
Output:
441
Time Complexity: O(M log M) for pre calculation where M is the maximum element and O(N) for iteration.
Space Complexity: O(M)
Note: As arr[i]2(N-1) can be really big, the answer can overflow, its preferable to use larger data-type and mod operations to conserve the answer.
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