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.
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 notbool 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 primestatic 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 codeint main(){ int n = 11; if (isStrongPrime(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program to check if given number is strong primeclass 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 primefrom math import sqrt# Utility function to check if a # number is prime or notdef 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 primedef 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 codeif __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 primeusing 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 notfunction 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 primefunction 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 notfunction 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 primefunction 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 codelet n = 11;if (isStrongPrime(n)) document.write("Yes");else document.write("No"); // This code is contributed by souravmahato348</script> |
Output:
Yes
Time Complexity: O(n1/2)
Auxiliary Space: O(1), since no extra space has been taken.
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