素数频率大于或等于 k 的数字

给定一个数组，找出在数组中出现素数次且频率最小为 k 的元素（频率 >= k）。

例子 ：

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 

方法 ：
1. 创建一个 Map，将数字作为 Key 和 value 作为其在输入数组中的出现次数。
2. 迭代 Map 键并查找与其键对应的值，返回对应的键
具有满足条件键的最小值的值是质数并且提供 >= 最小出现次数
作为输入。

C++

// C++ code to find number
// occurring prime number
// of times with frequency >= k
#include 
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 primeOccurences(int arr[], int k)
{
    unordered_map 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;
 
    primeOccurences(arr, k);
    return 0;
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

Java

// 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 primeOccurences(int[] arr, int k)
    {
        Map 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 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;
 
        primeOccurences(arr, k);
    }
}

Python3

# Python3 code to find number
# occurring prime number of
# times with frequency >= k
 
# Function to find number
# with prime occurrences
def primeOccurences(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
 
primeOccurences(arr, k)
 
 
# This code is contributed by Ansu Kumari.

C#

// 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 primeOccurences(int[] arr,
                                int k)
    {
        Dictionary map =
                   new Dictionary();
         
        // 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
                           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;
 
        primeOccurences(arr, k);
    }
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

Javascript

// 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 primeOccurences(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;
 
primeOccurences(arr, k);
 
// This code is contributed by
// gfgking
 

输出 ：

37
51

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