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Special prime numbers

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Given two numbers n and k, find whether there exist at least k Special prime numbers or not from 2 to n inclusively. 
A prime number is said to be Special prime number if it can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1. 
Note:- Two prime numbers are called neighboring if there are no other prime numbers between them. 
Examples: 
 

Input : n = 27, k = 2
Output : YES
In this sample the answer is YES
since at least two numbers are
Special 13(5 + 7 + 1) and
19(7 + 11 + 1).
Input : n = 45, k = 7
Output : NO
In this example, the Special
prime numbers are 13(5 + 7 + 1),
19(7 + 11 + 1), 31(13 + 17 + 1),
37(17 + 19 + 1), 43(19 + 23 + 1).
As the no. of Special prime
numbers from 2 to 45 is less than
k, the output is NO.


To solve this problem we need to find prime numbers in range [2..n]. So we us Sieve of Eratosthenes to generate all the prime numbers from 2 to n. Then, Take every pair of neighboring prime numbers and check if their sum increased by 1 is a prime number too. Count the number of these pairs, compare it to K and output the result. 
Below is the implementation of the above approach:- 

C++

// CPP program to check whether there
// exist at least k or not in range [2..n]
#include <bits/stdc++.h>
using namespace std;
 
vector<int> primes;
 
// Generating all the prime numbers
// from 2 to n.
void SieveofEratosthenes(int n)
{
    bool visited[n];
    for (int i = 2; i <= n + 1; i++)
        if (!visited[i]) {
            for (int j = i * i; j <= n + 1; j += i)
                visited[j] = true;
            primes.push_back(i);
        }
}
 
bool specialPrimeNumbers(int n, int k)
{
    SieveofEratosthenes(n);
    int count = 0;
    for (int i = 0; i < primes.size(); i++) {
        for (int j = 0; j < i - 1; j++) {
 
            // If a prime number is Special prime
            // number, then we increments the
            // value of k.
            if (primes[j] + primes[j + 1] + 1
                == primes[i]) {
                count++;
                break;
            }
        }
 
        // If at least k Special prime numbers
        // are present, then we return 1.
        // else we return 0 from outside of
        // the outer loop.
        if (count == k)
            return true;
    }
    return false;
}
 
// Driver function
int main()
{
    int n = 27, k = 2;
    if (specialPrimeNumbers(n, k))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
    return 0;
}

                    

Java

// Java program to check whether there
// exist at least k or not in range [2..n]
import java.util.*;
class GFG{
static ArrayList<Integer> primes = new ArrayList<Integer>();
// Generating all the prime numbers
// from 2 to n.
static void SieveofEratosthenes(int n)
{
    boolean[] visited=new boolean[n*n+2];
    for (int i = 2; i <= n + 1; i++)
        if (!visited[i]) {
            for (int j = i * i; j <= n + 1; j += i)
                visited[j] = true;
            primes.add(i);
        }
}
 
static boolean specialPrimeNumbers(int n, int k)
{
    SieveofEratosthenes(n);
    int count = 0;
    for (int i = 0; i < primes.size(); i++) {
        for (int j = 0; j < i - 1; j++) {
 
            // If a prime number is Special prime
            // number, then we increments the
            // value of k.
            if (primes.get(j) + primes.get(j + 1) + 1
                == primes.get(i)) {
                count++;
                break;
            }
        }
 
        // If at least k Special prime numbers
        // are present, then we return 1.
        // else we return 0 from outside of
        // the outer loop.
        if (count == k)
            return true;
    }
    return false;
}
 
// Driver function
public static void main(String[] args)
{
    int n = 27, k = 2;
    if (specialPrimeNumbers(n, k))
        System.out.println("YES");
    else
        System.out.println("NO");
}
}
// This code is contributed by mits

                    

Python3

# Python3 program to check whether there
# exist at least k or not in range [2..n]
primes = [];
 
# Generating all the prime numbers
# from 2 to n.
def SieveofEratosthenes(n):
 
    visited = [False] * (n + 2);
    for i in range(2, n + 2):
        if (visited[i] == False):
            for j in range(i * i, n + 2, i):
                visited[j] = True;
            primes.append(i);
 
def specialPrimeNumbers(n, k):
 
    SieveofEratosthenes(n);
    count = 0;
    for i in range(len(primes)):
        for j in range(i - 1):
 
            # If a prime number is Special
            # prime number, then we increments
            # the value of k.
            if (primes[j] +
                primes[j + 1] + 1 == primes[i]):
                count += 1;
                break;
 
        # If at least k Special prime numbers
        # are present, then we return 1.
        # else we return 0 from outside of
        # the outer loop.
        if (count == k):
            return True;
 
    return False;
 
# Driver Code
n = 27;
k = 2;
if (specialPrimeNumbers(n, k)):
    print("YES");
else:
    print("NO");
 
# This code is contributed by mits

                    

C#

// C# program to check whether there
// exist at least k or not in range [2..n]
using System;
using System.Collections;
 
class GFG{
static ArrayList primes = new ArrayList();
// Generating all the prime numbers
// from 2 to n.
static void SieveofEratosthenes(int n)
{
    bool[] visited=new bool[n*n+2];
    for (int i = 2; i <= n + 1; i++)
        if (!visited[i]) {
            for (int j = i * i; j <= n + 1; j += i)
                visited[j] = true;
            primes.Add(i);
        }
}
 
static bool specialPrimeNumbers(int n, int k)
{
    SieveofEratosthenes(n);
    int count = 0;
    for (int i = 0; i < primes.Count; i++) {
        for (int j = 0; j < i - 1; j++) {
 
            // If a prime number is Special prime
            // number, then we increments the
            // value of k.
            if ((int)primes[j] + (int)primes[j + 1] + 1
                == (int)primes[i]) {
                count++;
                break;
            }
        }
 
        // If at least k Special prime numbers
        // are present, then we return 1.
        // else we return 0 from outside of
        // the outer loop.
        if (count == k)
            return true;
    }
    return false;
}
 
// Driver function
public static void Main()
{
    int n = 27, k = 2;
    if (specialPrimeNumbers(n, k))
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
}
// This code is contributed by mits

                    

Javascript

<script>
 
    // Javascript program to check whether there
    // exist at least k or not in range [2..n]
     
    let primes = [];
   
    // Generating all the prime numbers
    // from 2 to n.
    function SieveofEratosthenes(n)
    {
        let visited = new Array(n);
        visited.fill(false);
        for (let i = 2; i <= n + 1; i++)
            if (!visited[i]) {
                for (let j = i * i; j <= n + 1; j += i)
                    visited[j] = true;
                primes.push(i);
            }
    }
 
    function specialPrimeNumbers(n, k)
    {
        SieveofEratosthenes(n);
        let count = 0;
        for (let i = 0; i < primes.length; i++) {
            for (let j = 0; j < i - 1; j++) {
 
                // If a prime number is Special prime
                // number, then we increments the
                // value of k.
                if (primes[j] + primes[j + 1] + 1
                    == primes[i]) {
                    count++;
                    break;
                }
            }
 
            // If at least k Special prime numbers
            // are present, then we return 1.
            // else we return 0 from outside of
            // the outer loop.
            if (count == k)
                return true;
        }
        return false;
    }
     
    let n = 27, k = 2;
    if (specialPrimeNumbers(n, k))
        document.write("YES");
    else
        document.write("NO");
 
</script>

                    

PHP

<?php
// PHP program to check whether there
// exist at least k or not in range [2..n]
$primes = array();
 
// Generating all the prime numbers
// from 2 to n.
function SieveofEratosthenes($n)
{
    global $primes;
    $visited = array_fill(0, $n, false);
    for ($i = 2; $i <= $n + 1; $i++)
        if (!$visited[$i])
        {
            for ($j = $i * $i;
                 $j <= $n + 1; $j += $i)
                $visited[$j] = true;
            array_push($primes, $i);
        }
}
 
function specialPrimeNumbers($n, $k)
{
    global $primes;
    SieveofEratosthenes($n);
    $count = 0;
    for ($i = 0; $i < count($primes); $i++)
    {
        for ($j = 0; $j < $i - 1; $j++)
        {
 
            // If a prime number is Special prime
            // number, then we increments the
            // value of k.
            if ($primes[$j] +
                $primes[$j + 1] + 1 == $primes[$i])
            {
                $count++;
                break;
            }
        }
 
        // If at least k Special prime numbers
        // are present, then we return 1.
        // else we return 0 from outside of
        // the outer loop.
        if ($count == $k)
            return true;
    }
    return false;
}
 
// Driver Code
$n = 27;
$k = 2;
if (specialPrimeNumbers($n, $k))
    echo "YES\n";
else
    echo "NO\n";
 
// This code is contributed by mits
?>

                    

Output:-

 YES

Approach#2: Using brute force

In this approach, we will iterate from 2 to n and check for each number if it is a special prime number or not. We will keep a count of the special prime numbers we have found so far and return “YES” if the count reaches k, otherwise “NO”.

Algorithm

1. Initialize count to 0.
2. For each number i from 2 to n:
a. Check if i is prime.
b. If i is prime, calculate the sum of i’s largest two prime factors and add 1 to it.
c. Check if the sum obtained in step b is a prime number or not.
d. If the sum is prime, increment the count.
e. If count is equal to k, return “YES”.
3. If we have not found k special prime numbers, return “NO”.

C++

// C++ code implementation
#include <cmath>
#include <iostream>
using namespace std;
 
bool is_prime(int x)
{
    if (x < 2) {
        return false;
    }
    for (int i = 2; i <= sqrt(x); i++) {
        if (x % i == 0) {
            return false;
        }
    }
    return true;
}
 
bool is_special_prime(int x)
{
    int s = 0;
    for (int i = 2; i <= x; i++) {
        if (is_prime(i)) {
            s += i;
        }
    }
    return is_prime(s);
}
 
string count_special_primes(int n, int k)
{
    int count = 0;
    for (int i = 2; i <= n; i++) {
        if (is_special_prime(i)) {
            count++;
        }
        if (count >= k) {
            return "YES";
        }
    }
    return "NO";
}
 
// Driver code
int main()
{
    int n = 27;
    int k = 2;
    cout << count_special_primes(n, k) << std::endl;
    return 0;
}

                    

Java

//Java code for the above approach
import java.util.*;
 
public class SpecialPrime {
 
    // Function to check if a number is prime
    public static boolean isPrime(int x) {
        if (x < 2) {
            return false;
        }
        for (int i = 2; i <= Math.sqrt(x); i++) {
            if (x % i == 0) {
                return false;
            }
        }
        return true;
    }
 
    // Function to check if a number is a special prime
    public static boolean isSpecialPrime(int x) {
        int s = 0;
        for (int i = 2; i <= x; i++) {
            if (isPrime(i)) {
                s += i;
            }
        }
        return isPrime(s);
    }
 
    // Function to count special primes up to 'n' and determine if there are at least 'k' of them
    public static String countSpecialPrimes(int n, int k) {
        int count = 0;
        for (int i = 2; i <= n; i++) {
            if (isSpecialPrime(i)) {
                count++;
            }
            if (count >= k) {
                return "YES";
            }
        }
        return "NO";
    }
 
    public static void main(String[] args) {
        int n = 27;
        int k = 2;
         
        // Call the function to count special primes and output the result
        System.out.println(countSpecialPrimes(n, k));
    }
}

                    

Python3

def count_special_primes(n, k):
    def is_prime(x):
        if x < 2:
            return False
        for i in range(2, int(x ** 0.5) + 1):
            if x % i == 0:
                return False
        return True
 
    def is_special_prime(x):
        s = 0
        for i in range(2, x + 1):
            if is_prime(i):
                s += i
        return is_prime(s)
 
    count = 0
    for i in range(2, n + 1):
        if is_special_prime(i):
            count += 1
        if count >= k:
            return "YES"
    return "NO"
n=27
k=2
print(count_special_primes(n, k))

                    

C#

using System;
 
class GFG
{
    static bool IsPrime(int x)
    {
        if (x < 2)
        {
            return false;
        }
        for (int i = 2; i <= Math.Sqrt(x); i++)
        {
            if (x % i == 0)
            {
                return false;
            }
        }
        return true;
    }
 
    static bool IsSpecialPrime(int x)
    {
        int s = 0;
        for (int i = 2; i <= x; i++)
        {
            if (IsPrime(i))
            {
                s += i;
            }
        }
        return IsPrime(s);
    }
 
    static string CountSpecialPrimes(int n, int k)
    {
        int count = 0;
        for (int i = 2; i <= n; i++)
        {
            if (IsSpecialPrime(i))
            {
                count++;
            }
            if (count >= k)
            {
                return "YES";
            }
        }
        return "NO";
    }
 
    static void Main(string[] args)
    {
        int n = 27;
        int k = 2;
        Console.WriteLine(CountSpecialPrimes(n, k));
    }
}

                    

Javascript

// Javascript code addition
function count_special_primes(n, k) {
    function is_prime(x) {
        if (x < 2) {
            return false;
        }
        for (let i = 2; i <= Math.sqrt(x); i++) {
            if (x % i === 0) {
                return false;
            }
        }
        return true;
    }
 
    function is_special_prime(x) {
        let s = 0;
        for (let i = 2; i <= x; i++) {
            if (is_prime(i)) {
                s += i;
            }
        }
        return is_prime(s);
    }
 
    let count = 0;
    for (let i = 2; i <= n; i++) {
        if (is_special_prime(i)) {
            count += 1;
        }
        if (count >= k) {
            return "YES";
        }
    }
    return "NO";
}
 
let n = 27;
let k = 2;
console.log(count_special_primes(n, k));
 
// The code is contributed by Anjali goel.

                    

Output
YES





Time Complexity: O(n^2loglogn) (iterating from 2 to n and then checking for each number if it is prime or not)

Auxiliary Space: O(1) (no extra space used) 



Last Updated : 19 Sep, 2023
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