# Check if N is Strong Prime

Given a positive integer N, the task is to check if N is a strong prime or not.
In number theory, a strong prime is a prime number that is greater than the arithmetic mean of nearest prime numbers i.e next and previous prime numbers.
First few strong prime numbers are 11, 17, 29, 37, 41, 59, 67, 71, …
A strong prime Pn can be represented as- where n is its index in the ordered set of prime numbers.

Examples:

Input: N = 11
Output: Yes
11 is 5th prime number, the arithmetic mean of 4th and 6th prime number i.e. 7 and 13 is 10.
11 is greater than 10 so 11 is a strong prime.

Input: N = 13
Output: No
13 is 6th prime number, the arithmetic mean of 5th (11) and 7th (17) is (11 + 17) / 2 = 14.
13 is smaller than 14 so 13 is not a strong prime.

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

Approach:

• If N is not a prime number or it is the first prime number i.e. 2 then print No.
• Else find the primes closest to N (one on the left and one on the right) and store their arithmetic mean in mean.
• If N > mean then print Yes.
• Else print No.

Below is the implementation of the above approach:

## C++

 `// C++ program to check if given number is strong prime ` `#include ` `using` `namespace` `std; ` ` `  `// Utility function to check ` `// if a number is prime or not ` `bool` `isPrime(``int` `n) ` `{ ` `    ``// Corner cases ` `    ``if` `(n <= 1) ` `        ``return` `false``; ` `    ``if` `(n <= 3) ` `        ``return` `true``; ` ` `  `    ``// This is checked so that we can skip ` `    ``// middle five numbers in below loop ` `    ``if` `(n % 2 == 0 || n % 3 == 0) ` `        ``return` `false``; ` ` `  `    ``for` `(``int` `i = 5; i * i <= n; i = i + 6) ` `        ``if` `(n % i == 0 || n % (i + 2) == 0) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Function that returns true if n is a strong prime ` `static` `bool` `isStrongPrime(``int` `n) ` `{ ` `    ``// If n is not a prime number or ` `    ``// n is the first prime then return false ` `    ``if` `(!isPrime(n) || n == 2) ` `        ``return` `false``; ` ` `  `    ``// Initialize previous_prime to n - 1 ` `    ``// and next_prime to n + 1 ` `    ``int` `previous_prime = n - 1; ` `    ``int` `next_prime = n + 1; ` ` `  `    ``// Find next prime number ` `    ``while` `(!isPrime(next_prime)) ` `        ``next_prime++; ` ` `  `    ``// Find previous prime number ` `    ``while` `(!isPrime(previous_prime)) ` `        ``previous_prime--; ` ` `  `    ``// Arithmetic mean ` `    ``int` `mean = (previous_prime + next_prime) / 2; ` ` `  `    ``// If n is a strong prime ` `    ``if` `(n > mean) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `n = 11; ` ` `  `    ``if` `(isStrongPrime(n)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to check if given number is strong prime ` `class` `GFG { ` ` `  `    ``// Utility function to check ` `    ``// if a number is prime or not ` `    ``static` `boolean` `isPrime(``int` `n) ` `    ``{ ` `        ``// Corner cases ` `        ``if` `(n <= ``1``) ` `            ``return` `false``; ` `        ``if` `(n <= ``3``) ` `            ``return` `true``; ` ` `  `        ``// This is checked so that we can skip ` `        ``// middle five numbers in below loop ` `        ``if` `(n % ``2` `== ``0` `|| n % ``3` `== ``0``) ` `            ``return` `false``; ` ` `  `        ``for` `(``int` `i = ``5``; i * i <= n; i = i + ``6``) ` `            ``if` `(n % i == ``0` `|| n % (i + ``2``) == ``0``) ` `                ``return` `false``; ` ` `  `        ``return` `true``; ` `    ``} ` ` `  `    ``// Function that returns true if n is a strong prime ` `    ``static` `boolean` `isStrongPrime(``int` `n) ` `    ``{ ` `        ``// If n is not a prime number or ` `        ``// n is the first prime then return false ` `        ``if` `(!isPrime(n) || n == ``2``) ` `            ``return` `false``; ` ` `  `        ``// Initialize previous_prime to n - 1 ` `        ``// and next_prime to n + 1 ` `        ``int` `previous_prime = n - ``1``; ` `        ``int` `next_prime = n + ``1``; ` ` `  `        ``// Find next prime number ` `        ``while` `(!isPrime(next_prime)) ` `            ``next_prime++; ` ` `  `        ``// Find previous prime number ` `        ``while` `(!isPrime(previous_prime)) ` `            ``previous_prime--; ` ` `  `        ``// Arithmetic mean ` `        ``int` `mean = (previous_prime + next_prime) / ``2``; ` ` `  `        ``// If n is a strong prime ` `        ``if` `(n > mean) ` `            ``return` `true``; ` `        ``else` `            ``return` `false``; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` ` `  `        ``int` `n = ``11``; ` ` `  `        ``if` `(isStrongPrime(n)) ` `            ``System.out.println(``"Yes"``); ` ` `  `        ``else` `            ``System.out.println(``"No"``); ` `    ``} ` `} `

## Python3

 `# Python 3 program to check if given  ` `# number is strong prime ` `from` `math ``import` `sqrt ` ` `  `# Utility function to check if a  ` `# number is prime or not ` `def` `isPrime(n): ` `     `  `    ``# Corner cases ` `    ``if` `(n <``=` `1``): ` `        ``return` `False` `    ``if` `(n <``=` `3``): ` `        ``return` `True` ` `  `    ``# This is checked so that we can skip ` `    ``# middle five numbers in below loop ` `    ``if` `(n ``%` `2` `=``=` `0` `or` `n ``%` `3` `=``=` `0``): ` `        ``return` `False` `     `  `    ``k ``=` `int``(sqrt(n)) ``+` `1` `    ``for` `i ``in` `range``(``5``, k, ``6``): ` `        ``if` `(n ``%` `i ``=``=` `0` `or` `n ``%` `(i ``+` `2``) ``=``=` `0``): ` `            ``return` `False` ` `  `    ``return` `True` ` `  `# Function that returns true if  ` `# n is a strong prime ` `def` `isStrongPrime(n): ` `     `  `    ``# If n is not a prime number or ` `    ``# n is the first prime then return false ` `    ``if` `(isPrime(n) ``=``=` `False` `or` `n ``=``=` `2``): ` `        ``return` `False` ` `  `    ``# Initialize previous_prime to n - 1 ` `    ``# and next_prime to n + 1 ` `    ``previous_prime ``=` `n ``-` `1` `    ``next_prime ``=` `n ``+` `1` ` `  `    ``# Find next prime number ` `    ``while` `(isPrime(next_prime) ``=``=` `False``): ` `        ``next_prime ``+``=` `1` ` `  `    ``# Find previous prime number ` `    ``while` `(isPrime(previous_prime) ``=``=` `False``): ` `        ``previous_prime ``-``=` `1` ` `  `    ``# Arithmetic mean ` `    ``mean ``=` `(previous_prime ``+` `next_prime) ``/` `2` ` `  `    ``# If n is a strong prime ` `    ``if` `(n > mean): ` `        ``return` `True` `    ``else``: ` `        ``return` `False` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `11` ` `  `    ``if` `(isStrongPrime(n)): ` `        ``print``(``"Yes"``) ` `    ``else``: ` `        ``print``(``"No"``) ` ` `  `# This code is contributed by ` `# Sanjit_prasad `

## C#

 `// C# program to check if a given number is strong prime ` `using` `System; ` `class` `GFG { ` ` `  `    ``// Utility function to check ` `    ``// if a number is prime or not ` `    ``static` `bool` `isPrime(``int` `n) ` `    ``{ ` `        ``// Corner cases ` `        ``if` `(n <= 1) ` `            ``return` `false``; ` `        ``if` `(n <= 3) ` `            ``return` `true``; ` ` `  `        ``// This is checked so that we can skip ` `        ``// middle five numbers in below loop ` `        ``if` `(n % 2 == 0 || n % 3 == 0) ` `            ``return` `false``; ` ` `  `        ``for` `(``int` `i = 5; i * i <= n; i = i + 6) ` `            ``if` `(n % i == 0 || n % (i + 2) == 0) ` `                ``return` `false``; ` ` `  `        ``return` `true``; ` `    ``} ` ` `  `    ``// Function that returns true if n is a strong prime ` `    ``static` `bool` `isStrongPrime(``int` `n) ` `    ``{ ` `        ``// If n is not a prime number or ` `        ``// n is the first prime then return false ` `        ``if` `(!isPrime(n) || n == 2) ` `            ``return` `false``; ` ` `  `        ``// Initialize previous_prime to n - 1 ` `        ``// and next_prime to n + 1 ` `        ``int` `previous_prime = n - 1; ` `        ``int` `next_prime = n + 1; ` ` `  `        ``// Find next prime number ` `        ``while` `(!isPrime(next_prime)) ` `            ``next_prime++; ` ` `  `        ``// Find previous prime number ` `        ``while` `(!isPrime(previous_prime)) ` `            ``previous_prime--; ` ` `  `        ``// Arithmetic mean ` `        ``int` `mean = (previous_prime + next_prime) / 2; ` ` `  `        ``// If n is a strong prime ` `        ``if` `(n > mean) ` `            ``return` `true``; ` `        ``else` `            ``return` `false``; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 11; ` ` `  `        ``if` `(isStrongPrime(n)) ` `            ``Console.WriteLine(``"Yes"``); ` `        ``else` `            ``Console.WriteLine(``"No"``); ` `    ``} ` `} `

## PHP

 ` ``\$mean``) ` `        ``return` `true; ` `    ``else` `        ``return` `false; ` `} ` ` `  `// Driver code ` `\$n` `= 11; ` ` `  `if` `(isStrongPrime(``\$n``)) ` `    ``echo` `(``"Yes"``); ` `else` `    ``echo``(``"No"``); ` ` `  `// This code is contributed  ` `// by Shivi_Aggarwal ` `?> `

Output:

```Yes
```

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