# Prime Subset Product Problem

• Difficulty Level : Hard
• Last Updated : 19 Dec, 2022

Given an array arr[] of N integers. The 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 subsets 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 time i.e.

$X^{2^{N-1}}&space;\mod&space;M$

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.

$X^{&space;(2^{N-1}&space;\mod&space;(M-1))&space;}&space;\mod&space;M$

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#includeusing 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 arrayint product(int A[], int n){     // Create Sieve to check whether a    // number is prime or not    int N = 100010;    int mod = 1000000007;    vector 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 codeint 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 approachclass 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 arraystatic 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 codepublic 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 arraydef 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 codeA = [3, 7]print(product(A))

## C#

 // C# implementation of the approachusing 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 arraystatic 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 codepublic 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



Output

441

Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(100010) ⇒ O(1), no extra space is required, so it is a constant.

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