Check if given number is Emirp Number or not

An Emirp Number (prime spelled backwards) is a prime number that results in a different prime when its decimal digits are reversed. This definition excludes the related palindromic primes. 
Examples :

Input : n = 13
Output : 13 is Emirp!
Explanation :
13 and 31 are both prime numbers. 
Thus, 13 is an Emirp number.

Input : n = 27
Output : 27 is not Emirp.

Objective : Input a number and find whether the number is an emirp number or not. 

Approach : Input a number and firstly check if its a prime number or not. If the number is a prime number, then we find the reverse of the original number and check the reversed number for being prime or not. If the reversed number is also prime, then the original number is an Emirp Number otherwise it is not.
Below is the implementation of above approach :. 

C++

// C++ program to check if given
// number is Emirp or not.
#include <iostream>
using namespace std;

// Returns true if n is prime.
// Else false.
bool isPrime(int n)
{
    // Corner case
    if (n <= 1)
        return false;

    // Check from 2 to n-1
    for (int i = 2; i < n; i++)
        if (n % i == 0)
            return false;

    return true;
}

// Function will check whether number
// is Emirp or not
bool isEmirp(int n)
{
    // Check if n is prime
    if (isPrime(n) == false)
        return false;

    // Find reverse of n
    int rev = 0;
    while (n != 0) {
        int d = n % 10;
        rev = rev * 10 + d;
        n /= 10;
    }

    // If both Original and Reverse are Prime,
    // then it is an Emirp number
    return isPrime(rev);
}

// Driver code
int main()
{
    int n = 13; // Input number
    if (isEmirp(n) == true)
        cout << "Yes";
    else
        cout << "No";
}

// This code is contributed by Anant Agarwal.

Java

// Java program to check if given number is
// Emirp or not.
import java.io.*;
class Emirp {
    // Returns true if n is prime. Else
    // false.
    public static boolean isPrime(int n)
    {
        // Corner case
        if (n <= 1)
            return false;

        // Check from 2 to n-1
        for (int i = 2; i < n; i++)
            if (n % i == 0)
                return false;

        return true;
    }

    // Function will check whether number
    // is Emirp or not
    public static boolean isEmirp(int n)
    {
        // Check if n is prime
        if (isPrime(n) == false)
            return false;

        // Find reverse of n
        int rev = 0;
        while (n != 0) {
            int d = n % 10;
            rev = rev * 10 + d;
            n /= 10;
        }

        // If both Original and Reverse are Prime,
        // then it is an Emirp number
        return isPrime(rev);
    }

    // Driver Function
    public static void main(String args[]) throws IOException
    {
        int n = 13; // Input number
        if (isEmirp(n) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python3

# Python3 code to check if
# given number is Emirp or not.

# Returns true if n is prime.
# Else false.
def isPrime( n ):

    # Corner case
    if n <= 1:
        return False

    # Check from 2 to n-1
    for i in range(2, n):
        if n % i == 0:
            return False

    return True

# Function will check whether
# number is Emirp or not
def isEmirp( n):

    # Check if n is prime
    n = int(n)
    if isPrime(n) == False:
        return False

        # Find reverse of n
    rev = 0
    while n != 0:
        d = n % 10
        rev = rev * 10 + d
        n = int(n / 10)


    # If both Original and Reverse
    # are Prime, then it is an
    # Emirp number
    return isPrime(rev)

# Driver Function
n = 13 # Input number
if isEmirp(n):
    print("Yes")
else:
    print("No")

# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to check if given
// number is Emirp or not.
using System;

class Emirp {
    // Returns true if n is prime
    // Else false.
    public static bool isPrime(int n)
    {
        // Corner case
        if (n <= 1)
            return false;

        // Check from 2 to n-1
        for (int i = 2; i < n; i++)
            if (n % i == 0)
                return false;

        return true;
    }

    // Function will check whether number
    // is Emirp or not
    public static bool isEmirp(int n)
    {
        // Check if n is prime
        if (isPrime(n) == false)
            return false;

        // Find reverse of n
        int rev = 0;
        while (n != 0) {
            int d = n % 10;
            rev = rev * 10 + d;
            n /= 10;
        }

        // If both Original and Reverse are Prime,
        // then it is an Emirp number
        return isPrime(rev);
    }

    // Driver Function
    public static void Main()
    {
        int n = 13; // Input number
        if (isEmirp(n) == true)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to check if given
// number is Emirp or not.

// Returns true if n
// is prime else false

function isPrime($n)
{
    // Corner case
    if ($n <= 1)
        return -1;

    // Check from 2 to n-1
    for ($i = 2; $i < $n; $i++)
        if ($n % $i == 0)
            return -1;

    return 1;
}

// Function will check
// whether number is
// Emirp or not
function isEmirp($n)
{
    // Check if n is prime
    if (isPrime($n) == -1)
        return -1;

    // Find reverse of n
    $rev = 0;
    while ($n != 0)
    {
        $d = $n % 10;
        $rev = $rev * 10 + $d;
        $n /= 10;
    }

    // If both Original and
    // Reverse are Prime,
    // then it is an Emirp number
    return isPrime($rev);
}

// Driver code
$n = 13;

if (isEmirp($n) ==-1)
    echo "Yes";
    else
    echo "No";

// This code is contributed by ajit
?>

Javascript

<script>

// javascript program to check if given number is
// Emirp or not.

    // Returns true if n is prime. Else
    // false.
    function isPrime(n)
    {
        // Corner case
        if (n <= 1)
            return false;

        // Check from 2 to n-1
        for (i = 2; i < n; i++)
            if (n % i == 0)
                return false;

        return true;
    }

    // Function will check whether number
    // is Emirp or not
    function isEmirp(n)
    {
        // Check if n is prime
        if (isPrime(n) == false)
            return false;

        // Find reverse of n
        var rev = 0;
        while (n != 0) {
            var d = n % 10;
            rev = rev * 10 + d;
            n = parseInt(n/10);
        }

        // If both Original and Reverse are Prime,
        // then it is an Emirp number
        return isPrime(rev);
    }

    // Driver Function

        var n = 13; // Input number
        if (isEmirp(n) == true)
            document.write("Yes");
        else
            document.write("No");

// This code contributed by Princi Singh
</script>

Output :

Yes

Time complexity: O(n)

Auxiliary Space:O(1)