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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``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 = prime = 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);``    ` `    ``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 = prime = 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

 ``
Output:
`441`

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