# 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<``int``, ``int``> 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<``int``,  ` `               ``int``> mp = ``new` `Dictionary<``int``, ` `                                        ``int``>(); ` `    ``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<``int``, ``int``> 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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