Special prime numbers

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 CPP implementation of the above approach:-

// 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;
}

Output:- YES


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