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Prime Subset Product Problem

  • Difficulty Level : Hard
  • Last Updated : 19 May, 2021

Given an array arr[] of N integers. Value of a subset of array A is defined as the product of all prime numbers in that subset. If there are no primes in the subset then the value of that subset is 1. The task is to calculate the product of values of all possible non-empty subsets of the given array modulus 100000007.
Examples: 
 

Input: arr[] = {3, 7} 
Output: 441 
val({3}) = 3 
val({7}) = 7 
val({3, 7}) = 3 * 7 = 21 
3 * 7 * 21 = 441
Input: arr[] = {1, 1, 1} 
Output:
 

 

Approach: Since it is known that a number occurs 2N – 1 times in all the subset of the given array of size N. So if a number X is prime then the contribution of X will be X * X * X * ….. * 2N – 1 times i.e.
 



Since 2N – 1 will also be a large number, it cannot be calculated directly. Fermat’s Theorem will be used to calculate the power here.
 

After that, the value of each element can be calculated easily.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
 
int power(int a, int b, int mod)
{
    int aa = 1;
    while(b)
    {
        if(b & 1)
        {
            aa = aa * a;
            aa %= mod;
        }
        a = a * a;
        a %= mod;
        b /= 2;
    }
    return aa;
}
 
// Function to return the prime subset
// product of the given array
int product(int A[], int n)
{
 
    // Create Sieve to check whether a
    // number is prime or not
    int N = 100010;
    int mod = 1000000007;
    vector<int> prime(N, 1);
    prime[0] = prime[1] = 0;
    int i = 2;
    while (i * i < N)
    {
        if (prime[i])
            for (int j = 2 * i;
                     j <= N;j += i)
                prime[j] = 0;
 
        i += 1;
    }
 
    // Length of the array
    // Calculating 2^(n-1) % mod
    int t = power(2, n - 1, mod - 1);
 
    int ans = 1;
 
    for (int j = 0; j < n; j++)
    {
        int i = A[j];
 
        // If element is prime then add
        // its contribution in the result
        if( prime[i])
        {
            ans *= power(i, t, mod);
            ans %= mod;
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    int A[] = {3, 7};
     
    int n = sizeof(A) / sizeof(A[0]);
     
    printf("%d", product(A, n));
}
 
// This code is contributed by Mohit Kumar

Java




// Java implementation of the approach
class GFG
{
static int power(int a, int b, int mod)
{
    int aa = 1;
    while(b > 0)
    {
        if(b % 2 == 1)
        {
            aa = aa * a;
            aa %= mod;
        }
        a = a * a;
        a %= mod;
        b /= 2;
    }
    return aa;
}
 
// Function to return the prime subset
// product of the given array
static int product(int A[], int n)
{
 
    // Create Sieve to check whether a
    // number is prime or not
    int N = 100010;
    int mod = 1000000007;
    int []prime = new int[N];
    for (int j = 0; j < N; j++)
    {
        prime[j] = 1;
    }
     
    prime[0] = prime[1] = 0;
    int i = 2;
    while (i * i < N)
    {
        if (prime[i] == 1)
            for (int j = 2 * i;
                    j < N;j += i)
                prime[j] = 0;
 
        i += 1;
    }
 
    // Length of the array
    // Calculating 2^(n-1) % mod
    int t = power(2, n - 1, mod - 1);
 
    int ans = 1;
 
    for (int j = 0; j < n; j++)
    {
        i = A[j];
 
        // If element is prime then add
        // its contribution in the result
        if( prime[i] == 1)
        {
            ans *= power(i, t, mod);
            ans %= mod;
        }
    }
    return ans;
}
 
// Driver code
public static void main (String[] args)
{
    int A[] = {3, 7};
     
    int n = A.length;
     
    System.out.printf("%d", product(A, n));
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 implementation of the approach
 
# Function to return the prime subset
# product of the given array
def product(A):
     
    # Create Sieve to check whether a
    # number is prime or not
    N = 100010
    mod = 1000000007
    prime = [1] * N
    prime[0] = prime[1] = 0
    i = 2
    while i * i < N:
        if prime[i]:
            for j in range(i * i, N, i):
                prime[j] = 0
         
        i += 1
     
    # Length of the array
    n = len(A)
     
    # Calculating 2^(n-1) % mod
    t = pow(2, n-1, mod-1)
     
    ans = 1
     
    for i in A:
         
        # If element is prime then add
        # its contribution in the result
        if prime[i]:
            ans *= pow(i, t, mod)
            ans %= mod
             
    return ans
     
# Driver code
A = [3, 7]
print(product(A))

C#




// C# implementation of the approach
using System;
 
class GFG
{
static int power(int a, int b, int mod)
{
    int aa = 1;
    while(b > 0)
    {
        if(b % 2 == 1)
        {
            aa = aa * a;
            aa %= mod;
        }
        a = a * a;
        a %= mod;
        b /= 2;
    }
    return aa;
}
 
// Function to return the prime subset
// product of the given array
static int product(int []A, int n)
{
 
    // Create Sieve to check whether a
    // number is prime or not
    int N = 100010;
    int mod = 1000000007;
    int []prime = new int[N];
    for (int j = 0; j < N; j++)
    {
        prime[j] = 1;
    }
     
    prime[0] = prime[1] = 0;
    int i = 2;
    while (i * i < N)
    {
        if (prime[i] == 1)
            for (int j = 2 * i;
                     j < N; j += i)
                prime[j] = 0;
 
        i += 1;
    }
 
    // Length of the array
    // Calculating 2^(n-1) % mod
    int t = power(2, n - 1, mod - 1);
 
    int ans = 1;
 
    for (int j = 0; j < n; j++)
    {
        i = A[j];
 
        // If element is prime then add
        // its contribution in the result
        if( prime[i] == 1)
        {
            ans *= power(i, t, mod);
            ans %= mod;
        }
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int []A = {3, 7};
     
    int n = A.Length;
     
    Console.Write("{0}", product(A, n));
}
}
     
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// JavaScript implementation of the approach
 
function power(a, b, mod) {
    let aa = 1;
    while (b) {
        if (b & 1) {
            aa = aa * a;
            aa %= mod;
        }
        a = a * a;
        a %= mod;
        b = Math.floor(b / 2);
    }
    return aa;
}
 
// Function to return the prime subset
// product of the given array
function product(A, n) {
 
    // Create Sieve to check whether a
    // number is prime or not
    let N = 100010;
    let mod = 1000000007;
    let prime = new Array(N).fill(1);
    prime[0] = prime[1] = 0;
    let i = 2;
    while (i * i < N) {
        if (prime[i])
            for (let j = 2 * i;
                j <= N; j += i)
                prime[j] = 0;
 
        i += 1;
    }
 
    // Length of the array
    // Calculating 2^(n-1) % mod
    let t = power(2, n - 1, mod - 1);
 
    let ans = 1;
 
    for (let j = 0; j < n; j++) {
        let i = A[j];
 
        // If element is prime then add
        // its contribution in the result
        if (prime[i]) {
            ans *= power(i, t, mod);
            ans %= mod;
        }
    }
    return ans;
}
 
// Driver code
 
let A = [3, 7];
 
let n = A.length;
 
document.write(product(A, n));
 
// This code is contributed by Saurabh Jaiswal
 
</script>
Output: 
441

 

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