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++ 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 not void 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 subset int 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 Code int main()
{ int a[] = { 3, 7 };
int n = sizeof (a) / sizeof (a[0]);
cout << sumOfSubset(a, n);
} |
// 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 not static 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 subset static 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 Code public 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 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 not def 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 subset def 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 Code a = [ 3 , 7 ]
n = len (a)
print (SumOfSubset(a, n))
# This code is contributed # by Mohit Kumar |
// 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 not static 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 subset static 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 Code public static void Main(String[] args)
{ int []a = { 3, 7 };
int n = a.Length;
Console.WriteLine(sumOfSubset(a, n));
} } // This code is contributed by PrinciRaj1992 |
<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> |
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.