# 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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 primes = ``new` `ArrayList(); ` `// 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 `

## PHP

 ` `

Output:-

``` YES
```

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