Números con frecuencias primas mayores o iguales a k

Dada una array, busque elementos que aparezcan un número primo de veces en la array con una frecuencia mínima k (frecuencia >= k).

Ejemplos:  

Input : int[] arr = { 11, 11, 11, 23, 11, 37, 51, 
                      37, 37, 51, 51, 51, 51 };
        k = 2
Output : 37, 51
Explanation :
11's count is 4, 23 count 1, 37 count 3, 51 count 5. 
37 and 51 are two number that appear prime number of
time and frequencies greater than or equal to k.

Input : int[] arr = { 11, 22, 33 } min Occurrence = 1
Output : -1
None of the count is prime number of times 

Enfoque: 
1. Cree un mapa que contenga el número como clave y el valor como sus ocurrencias en la array de entrada. 
2. Repita las claves del mapa y busque los valores correspondientes a sus claves, devuelva la clave que 
tiene el valor mínimo que cumple la condición. El valor de la clave es un número primo y >= ocurrencia mínima proporcionada 
como entrada. 
 

// C++ code to find number
// occurring prime number
// of times with frequency >= k
#include <bits/stdc++.h>
using namespace std;
 
// Check if the number of
// occurrences are primes
// or not
bool isPrime(int n)
{
    // Corner case
    if (n <= 1) return false;
 
    // Check from 2 to n-1
    for (int i = 2; i < n; i++)
        if (n % i == 0)
            return false;
 
    return true;
}
 
// Function to find number
// with prime occurrences
void primeOccurrences(int arr[], int k)
{
    unordered_map<int, int> map;
     
    // Insert values and
    // their frequencies
    for (int i = 0; i < 12; i++)
        map[arr[i]]++;
 
    // Traverse map and find
    // elements with prime
    // frequencies and frequency
    // at least k
    for (auto x : map)
    {
        if (isPrime(x.second) &&
                    x.second >= k)
            cout << x.first << endl;
    }
}
 
// Driver code
int main()
{
    int arr[] = {11, 11, 11, 23,
                 11, 37, 37, 51,
                 51, 51, 51, 51};
    int k = 2;
 
    primeOccurrences(arr, k);
    return 0;
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

// Java code to find number occurring prime
// number of times with frequency >= k
import java.util.*;
 
public class PrimeNumber {
 
    // Function to find number with prime occurrences
    static void primeOccurrences(int[] arr, int k)
    {
        Map<Integer, Integer> map = new HashMap<>();
 
        // Insert values and their frequencies
        for (int i = 0; i < arr.length; i++) {
            int val = arr[i];
 
            int freq;
            if (map.containsKey(val)) {
                freq = map.get(val);
                freq++;
            }
            else
                freq = 1;
            map.put(val, freq);
        }
 
        // Traverse map and find elements with
        // prime frequencies and frequency at
        // least k
        for (Map.Entry<Integer, Integer> entry :
                               map.entrySet()) {
            int value = entry.getValue();
            if (isPrime(value) && value >= k)
                System.out.println(entry.getKey());
        }
    }
 
    // Check if the number of occurrences
    // are primes or not
    private static boolean isPrime(int n)
    {
 
        if ((n > 2 && n % 2 == 0) || n == 1)
            return false;       
 
        for (int i = 3; i <= (int)Math.sqrt(n);
             i += 2) {
            if (n % i == 0)
                return false;           
        }
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 11, 11, 11, 23, 11, 37,
                      37, 51, 51, 51, 51, 51 };
        int k = 2;
 
        primeOccurrences(arr, k);
    }
}

# Python3 code to find number
# occurring prime number of
# times with frequency >= k
 
# Function to find number
# with prime occurrences
def primeOccurrences(arr, k):
    map = {}
 
    # Insert values and their frequencies
    for val in arr:
        freq = 0
         
        if val in map :
            freq = map[val]
            freq += 1
             
        else :
            freq = 1
        map[val] = freq
 
    # Traverse map and find elements
    # with prime frequencies and
    # frequency at least k
    for entry in map :
        value = map[entry]
         
        if isPrime(value) and value >= k:
            print(entry)
 
# Check if the number of occurrences
# are primes or not
def isPrime(n):
 
    if (n > 2 and not n % 2) or n == 1:
        return False    
 
    for i in range(3, int(n**0.5 + 1), 2):
        if not n % i:
            return False
             
    return True
 
 
# Driver code
 
arr = [ 11, 11, 11, 23, 11, 37,
        37, 51, 51, 51, 51, 51 ]
k = 2
 
primeOccurrences(arr, k)
 
 
# This code is contributed by Ansu Kumari.

// C# code to find number
// occurring prime number
// of times with frequency >= k
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to find number
    // with prime occurrences
    static void primeOccurrences(int[] arr,
                                int k)
    {
        Dictionary<int, int> map =
                   new Dictionary<int, int>();
         
        // Insert values and
        // their frequencies
        for (int i = 0; i < arr.Length; i++)
        {
            int val = arr[i];
 
            int freq;
            if (map.ContainsKey(val))
            {
                freq = map[val];
                freq++;
                map.Remove(val);
            }
            else
                freq = 1;
            map.Add(val, freq);
        }
 
        // Traverse map and find elements
        // with prime frequencies and
        // frequency atleast k
        foreach (KeyValuePair<int, int>
                           pair in map)
        {
            int value = pair.Value;
            if (isPrime(value) &&
                        value >= k)
                Console.WriteLine(pair.Key);
        }
    }
 
    // Check if the number
    // of occurrences
    // are primes or not
    static bool isPrime(int n)
    {
        if ((n > 2 &&
             n % 2 == 0) || n == 1)
            return false;    
 
        for (int i = 3;
                 i <= (int)Math.Sqrt(n);
                 i += 2)
            {
                if (n % i == 0)
                    return false;        
            }
        return true;
    }
 
    // Driver code
    static void Main()
    {
        int[] arr = new int[]{11, 11, 11, 23, 11, 37,
                              37, 51, 51, 51, 51, 51};
        int k = 2;
 
        primeOccurrences(arr, k);
    }
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

<script>
// Javascript code to find number
// occurring prime number
// of times with frequency >= k
 
// Check if the number of
// occurrences are primes
// or not
function isPrime(n) {
    // Corner case
    if (n <= 1) return false;
 
    // Check from 2 to n-1
    for (let i = 2; i < n; i++)
        if (n % i == 0)
            return false;
 
    return true;
}
 
// Function to find number
// with prime occurrences
function primeOccurrences(arr, k) {
    let map = new Map();
 
    // Insert values and
    // their frequencies
    for (let i = 0; i < arr.length; i++) {
        let val = arr[i];
 
        let freq;
        if (map.has(val)) {
            freq = map.get(val);
            freq++;
        }
        else
            freq = 1;
        map.set(val, freq);
    }
 
    // Traverse map and find
    // elements with prime
    // frequencies and frequency
    // at least k
    for (let x of map) {
        if (isPrime(x[1]) &&
            x[1] >= k)
            document.write(x[0] + "<br>");
    }
}
 
// Driver code
 
let arr = [11, 11, 11, 23,
           11, 37, 37, 51,
              51, 51, 51, 51];
let k = 2;
 
primeOccurrences(arr, k);
 
// This code is contributed by
// gfgking
 
</script>

37
51

Complejidad de tiempo: O(n), donde n es el número de elementos en la array.
Espacio auxiliar: O(n), donde n es el número de elementos de la array.