# Product of elements in an array having prime frequency

Given an array arr[] of N elements, the task is to find the product of the elements which have prime frequencies in the array. Since, the product can be large so print the product modulo 109 + 7. Note that 1 is neither prime nor composite.

Examples:

Input: arr[] = {5, 4, 6, 5, 4, 6}
Output: 120
All the elements appear 2 times which is a prime
So, 5 * 4 * 6 = 120

Input: arr[] = {1, 2, 3, 3, 2, 3, 2, 3, 3}
Output: 6
Only 2 and 3 appears prime number of times i.e. 3 and 5 respectively.
So, 2 * 3 = 6

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Traverse the array and store the frequencies of all the elements in a map.
• Build Sieve of Eratosthenes which will be used to test the primality of a number in O(1) time.
• Calculate the product of elements having prime frequency using the Sieve array calculated in the previous step.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std;    #define MOD 1000000007    // Function to create Sieve to check primes void SieveOfEratosthenes(bool prime[], int p_size) {     // False here indicates     // that it is not prime     prime[0] = false;     prime[1] = false;        for (int p = 2; p * p <= p_size; p++) {            // If prime[p] is not changed,         // then it is a prime         if (prime[p]) {                // Update all multiples of p,             // set them to non-prime             for (int i = p * 2; i <= p_size; i += p)                 prime[i] = false;         }     } }    // Function to return the product of elements // in an array having prime frequency int productPrimeFreq(int arr[], int n) {     bool prime[n + 1];     memset(prime, true, sizeof(prime));        SieveOfEratosthenes(prime, n + 1);        int i, j;        // Map is used to store     // element frequencies     unordered_map m;     for (i = 0; i < n; i++)         m[arr[i]]++;        long product = 1;        // Traverse the map using iterators     for (auto it = m.begin(); it != m.end(); it++) {            // Count the number of elements         // having prime frequencies         if (prime[it->second]) {             product *= (it->first % MOD);             product %= MOD;         }     }        return (int)(product); }    // Driver code int main() {     int arr[] = { 5, 4, 6, 5, 4, 6 };     int n = sizeof(arr) / sizeof(arr[0]);        cout << productPrimeFreq(arr, n);        return 0; }

## Java

 // Java implementation of the approach import java.util.*;        class GFG { static int MOD = 1000000007;    // Function to create Sieve to check primes static void SieveOfEratosthenes(boolean prime[],                                  int p_size) {     // False here indicates     // that it is not prime     prime[0] = false;     prime[1] = false;        for (int p = 2; p * p <= p_size; p++)      {            // If prime[p] is not changed,         // then it is a prime         if (prime[p])         {                // Update all multiples of p,             // set them to non-prime             for (int i = p * 2;                       i <= p_size; i += p)                 prime[i] = false;         }     } }    // Function to return the product of elements // in an array having prime frequency static int productPrimeFreq(int arr[], int n) {     boolean []prime = new boolean[n + 1];     for (int i = 0; i < n; i++)         prime[i] = true;        SieveOfEratosthenes(prime, n + 1);        int i, j;        // Map is used to store     // element frequencies     HashMap mp = new HashMap();        for (i = 0 ; i < n; i++)     {         if(mp.containsKey(arr[i]))         {             mp.put(arr[i], mp.get(arr[i]) + 1);         }         else         {             mp.put(arr[i], 1);         }     }     long product = 1;        // Traverse the map using iterators     for (Map.Entry it : mp.entrySet())      {            // Count the number of elements         // having prime frequencies         if (prime[it.getValue()])          {             product *= (it.getKey() % MOD);             product %= MOD;         }     }     return (int)(product); }    // Driver code static public void main (String []arg) {     int arr[] = { 5, 4, 6, 5, 4, 6 };     int n = arr.length;        System.out.println(productPrimeFreq(arr, n)); } }    // This code is contributed by Rajput-Ji

## Python3

 # Python3 implementation of the approach MOD = 1000000007    # Function to create Sieve to check primes def SieveOfEratosthenes(prime, p_size):            # False here indicates     # that it is not prime     prime[0] = False     prime[1] = False        for p in range(2, p_size):            # If prime[p] is not changed,         # then it is a prime         if (prime[p]):                # Update all multiples of p,             # set them to non-prime             for i in range(2 * p, p_size, p):                 prime[i] = False    # Function to return the product of elements # in an array having prime frequency def productPrimeFreq(arr, n):     prime = [True for i in range(n + 1)]        SieveOfEratosthenes(prime, n + 1)        i, j = 0, 0        # Map is used to store     # element frequencies     m = dict()     for i in range(n):         m[arr[i]] = m.get(arr[i], 0) + 1        product = 1        # Traverse the map using iterators     for it in m:            # Count the number of elements         # having prime frequencies         if (prime[m[it]]):             product *= it % MOD             product %= MOD        return product    # Driver code arr = [5, 4, 6, 5, 4, 6] n = len(arr)    print(productPrimeFreq(arr, n))    # This code is contributed by Mohit Kumar

## C#

 // C# implementation for above approach using System; using System.Collections.Generic;     class GFG { static int MOD = 1000000007;    // Function to create Sieve to check primes static void SieveOfEratosthenes(bool []prime,                                  int p_size) {     // False here indicates     // that it is not prime     prime[0] = false;     prime[1] = false;        for (int p = 2; p * p <= p_size; p++)      {            // If prime[p] is not changed,         // then it is a prime         if (prime[p])         {                // Update all multiples of p,             // set them to non-prime             for (int i = p * 2;                       i <= p_size; i += p)                 prime[i] = false;         }     } }    // Function to return the product of elements // in an array having prime frequency static int productPrimeFreq(int []arr, int n) {     bool []prime = new bool[n + 1];     int i;     for (i = 0; i < n; i++)         prime[i] = true;        SieveOfEratosthenes(prime, n + 1);        // Map is used to store     // element frequencies     Dictionary mp = new Dictionary();     for (i = 0 ; i < n; i++)     {         if(mp.ContainsKey(arr[i]))         {             var val = mp[arr[i]];             mp.Remove(arr[i]);             mp.Add(arr[i], val + 1);          }         else         {             mp.Add(arr[i], 1);         }     }     long product = 1;        // Traverse the map using iterators     foreach(KeyValuePair it in mp)     {            // Count the number of elements         // having prime frequencies         if (prime[it.Value])          {             product *= (it.Key % MOD);             product %= MOD;         }     }     return (int)(product); }    // Driver code static public void Main (String []arg) {     int []arr = { 5, 4, 6, 5, 4, 6 };     int n = arr.Length;        Console.WriteLine(productPrimeFreq(arr, n)); } }    // This code is contributed by Princi Singh

Output:

120

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