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Check if N is Strong Prime

  • Difficulty Level : Medium
  • Last Updated : 08 Apr, 2021

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- 
 

Strong prime

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. 
 

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 <bits/stdc++.h>
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




<?php
// PHP program to check if given number
// is strong isPrime
 
// Utility function to check if a
// number is prime or not
function 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 || $n % 3 == 0)
        return false;
 
    for ($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
function isStrongPrime($n)
{
    // If n is not a prime number or
    // n is the first prime then return false
    if (!isPrime($n) || $n == 2)
        return false;
 
    // Initialize previous_prime to n - 1
    // and next_prime to n + 1
    $previous_prime = $n - 1;
    $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
    $mean = ($previous_prime +
             $next_prime) / 2;
 
    // If n is a strong prime
    if ($n > $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
?>

Javascript




<script>
 
// Javascript program to check if
// given number is strong prime
 
// Utility function to check
// if a number is prime or not
function 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 || n % 3 == 0)
        return false;
 
    for(let 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
function isStrongPrime(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
    let previous_prime = n - 1;
    let 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
    let mean = parseInt((previous_prime +
                         next_prime) / 2);
 
    // If n is a strong prime
    if (n > mean)
        return true;
    else
        return false;
}
 
// Driver code
let n = 11;
 
if (isStrongPrime(n))
    document.write("Yes");
else
    document.write("No");
     
// This code is contributed by souravmahato348
 
</script>
Output: 
Yes

 

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