素数频率大于或等于k的数

给定一个数组,查找在数组中出现素数次且频率最小为k(频率>=k)的元素。

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例如:

Input : int[] arr = { 11, 11, 11, 23, 11, 37, 51,                       37, 37, 51, 51, 51, 51 };        k = 2Output : 37, 51Explanation :11's count is 4, 23 count 1, 37 count 3, 51 count 5. 37 and 51 are two number that appear prime number oftime and frequencies greater than or equal to k.Input : int[] arr = { 11, 22, 33 } min Occurrence = 1Output : -1None of the count is prime number of times 

方法: 1.创建一个映射,将数字作为键,将值作为其在输入数组中的引用。 2.迭代映射键并查找对应于其键的值,返回 具有满足条件的最小值键的值为素数且>=min 作为输入。

C++

// 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 primeOccurences( 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;
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<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 ;
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< 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;
primeOccurences(arr, k);
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)


Javascript

<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 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
</script>


输出:

3751

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