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Check if given number is Emirp Number or not

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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 the 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.

                    

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>

                    

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
?>

                    

Output : 
 

Yes

Time complexity: O(n)

Auxiliary Space:O(1)
 

Approach#2: Using recursion

Check if the given number is prime or not If it is prime, reverse the number and check if the reversed number is also prime or not If both the original and reversed numbers are prime, then the given number is an Emirp number

Algorithm

1. Define a function is_prime() to check if a number is prime or not
2. Define a function reverse_number() to reverse a given number
3. Define a function is_emirp() which takes a number as input
4. Check if the given number is prime or not using is_prime() function
5. If it is not prime, return “Not Emirp”
6. Reverse the given number using reverse_number() function
7. Check if the reversed number is prime using is_prime() function
8. If both the original and reversed numbers are prime, return “Emirp”, else return “Not Emirp”

C++

#include <cmath>
#include <iostream>
#include <string>
 
using namespace std;
 
// Function to check if a number is prime
bool is_prime(int num)
{
    if (num < 2) {
        return false;
    }
    for (int i = 2; i <= sqrt(num); i++) {
        if (num % i == 0) {
            return false;
        }
    }
    return true;
}
 
// Function to reverse a number
int reverse_number(int num)
{
    string str_num = to_string(num);
    string rev_str_num = "";
    for (int i = str_num.length() - 1; i >= 0; i--) {
        rev_str_num += str_num[i];
    }
    return stoi(rev_str_num);
}
 
// Function to check if a number is an emirp
string is_emirp(int num)
{
    if (!is_prime(num)) {
        return "Not Emirp";
    }
    int rev_num = reverse_number(num);
    if (is_prime(rev_num) && num != rev_num) {
        return "Emirp";
    }
    else {
        return "Not Emirp";
    }
}
 
int main()
{
    int num = 27;
    cout << is_emirp(num) << endl;
}

                    

Java

import java.lang.Math;
import java.util.Scanner;
 
public class Emirp {
    // Function to check if a number is prime
    public static boolean isPrime(int num) {
        if (num < 2) {
            return false;
        }
        for (int i = 2; i <= Math.sqrt(num); i++) {
            if (num % i == 0) {
                return false;
            }
        }
        return true;
    }
 
    // Function to reverse a number
    public static int reverseNumber(int num) {
        String strNum = Integer.toString(num);
        String revStrNum = "";
        for (int i = strNum.length() - 1; i >= 0; i--) {
            revStrNum += strNum.charAt(i);
        }
        return Integer.parseInt(revStrNum);
    }
 
    // Function to check if a number is an emirp
    public static String isEmirp(int num) {
        if (!isPrime(num)) {
            return "Not Emirp";
        }
        int revNum = reverseNumber(num);
        if (isPrime(revNum) && num != revNum) {
            return "Emirp";
        } else {
            return "Not Emirp";
        }
    }
 
    public static void main(String[] args) {
        int num = 27;
        System.out.println(isEmirp(num));
    }
}

                    

Python3

def is_prime(num):
    if num < 2:
        return False
    for i in range(2, int(num**0.5)+1):
        if num % i == 0:
            return False
    return True
 
 
def reverse_number(num):
    return int(str(num)[::-1])
 
 
def is_emirp(num):
    if not is_prime(num):
        return "Not Emirp"
    rev_num = reverse_number(num)
    if is_prime(rev_num) and num != rev_num:
        return "Emirp"
    else:
        return "Not Emirp"
 
 
num = 27
print(is_emirp(num))

                    

C#

using System;
 
namespace Emirp
{
    class GFG
    {
        // Function to check if a number is prime
        static bool IsPrime(int num)
        {
            if (num < 2)
            {
                return false;
            }
            for (int i = 2; i <= Math.Sqrt(num); i++)
            {
                if (num % i == 0)
                {
                    return false;
                }
            }
            return true;
        }
 
        // Function to reverse a number
        static int ReverseNumber(int num)
        {
            string strNum = num.ToString();
            char[] charArray = strNum.ToCharArray();
            Array.Reverse(charArray);
            string revStrNum = new string(charArray);
            return int.Parse(revStrNum);
        }
 
        // Function to check if a number is an emirp
        static string IsEmirp(int num)
        {
            if (!IsPrime(num))
            {
                return "Not Emirp";
            }
            int revNum = ReverseNumber(num);
            if (IsPrime(revNum) && num != revNum)
            {
                return "Emirp";
            }
            else
            {
                return "Not Emirp";
            }
        }
 
        static void Main(string[] args)
        {
            int num = 27;
            Console.WriteLine(IsEmirp(num));
        }
    }
}

                    

Javascript

function isPrime(num) {
  if (num < 2) {
    return false;
  }
  for (let i = 2; i <= Math.sqrt(num); i++) {
    if (num % i === 0) {
      return false;
    }
  }
  return true;
}
 
function reverseNumber(num) {
  return parseInt(num.toString().split('').reverse().join(''));
}
 
function isEmirp(num) {
  if (!isPrime(num)) {
    return "Not Emirp";
  }
  const revNum = reverseNumber(num);
  if (isPrime(revNum) && num !== revNum) {
    return "Emirp";
  } else {
    return "Not Emirp";
  }
}
 
const num = 27;
console.log(isEmirp(num));

                    

Output
Not Emirp

Time Complexity: O(sqrt(n)), n is the input number for which we are checking if it is an Emirp number or not.
Auxiliary Space: O(log n), n is the input number for which we are checking if it is an Emirp number or not.



Last Updated : 02 Aug, 2023
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