Largest palindromic prime in an array

Given an array arr[] of integers, the task is to print the largest palindromic prime from the array. If no element from the array is a palindromic prime then print -1.

Examples:

Input: arr[] = {11, 5, 121, 7, 89}
Output: 11
11, 5 and 7 are the only primes from the array which are palindromes.
11 is the maximum among them.

Input: arr[] = {2, 4, 6, 8, 10}
Output: 2

A simple approach is to go through every array element, check if it is prime and check if it is palindrome. If yes, the update the result if it is greater than current result also.

Efficient approach for large number of elements:

  • Use Sieve of Eratosthenes to calculate whether a number is prime or not upto the maximum element from the array.
  • Now, initialize a variable currentMax = -1 and start traversing the array arr[].
  • For every i, if arr[i] is prime as well as palindrome and arr[i] > currentMax then update currentMax = arr[i].
  • Print currentMax in the end.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if n is a palindrome
bool isPal(int n)
{
    // Find the appropriate divisor
    // to extract the leading digit
    int divisor = 1;
    while (n / divisor >= 10)
        divisor *= 10;
  
    while (n != 0) {
        int leading = n / divisor;
        int trailing = n % 10;
  
        // If first and last digit
        // not same return false
        if (leading != trailing)
            return false;
  
        // Removing the leading and trailing
        // digit from number
        n = (n % divisor) / 10;
  
        // Reducing divisor by a factor
        // of 2 as 2 digits are dropped
        divisor = divisor / 100;
    }
    return true;
}
  
// Function to return the largest
// palindromic prime present in the array
int maxPalindromicPrime(int arr[], int n)
{
    int maxElement = *max_element(arr, arr + n);
  
    // Create a boolean array "prime[0..n]" and
    // initialize all entries it as true. A value
    // in prime[i] will finally be false if i is
    // Not a prime, else true.
    bool prime[maxElement + 1];
    memset(prime, true, sizeof(prime));
  
    // 0 and 1 are not primes
    prime[0] = prime[1] = false;
    for (int p = 2; p * p <= maxElement; p++) {
  
        // If prime[p] is not changed, then it is
        // a prime
        if (prime[p] == true) {
  
            // Update all multiples of p
            for (int i = p * 2; i <= maxElement; i += p)
                prime[i] = false;
        }
    }
  
    int currentMax = -1;
    for (int i = 0; i < n; i++)
  
        // If arr[i] is prime as well as palindrome
        if (prime[arr[i]] && isPal(arr[i]))
            currentMax = max(currentMax, arr[i]);
  
    return currentMax;
}
  
// Driver Program
int main()
{
    int arr[] = { 11, 5, 121, 7, 89 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << maxPalindromicPrime(arr, n);
    return 0;
}

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Java














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// Java implementation of the above approach  


  


import java.util.Arrays; 


public class GFG{ 


   


    // Function that returns true if n is a palindrome  


    static boolean isPal(int n)  


    {  


        // Find the appropriate divisor  


        // to extract the leading digit  


        int divisor = 1;  


        while (n / divisor >= 10)  


            divisor *= 10;  


       


        while (n != 0) {  


            int leading = n / divisor;  


            int trailing = n % 10;  


       


            // If first and last digit  


            // not same return false  


            if (leading != trailing)  


                return false;  


       


            // Removing the leading and trailing  


            // digit from number  


            n = (n % divisor) / 10;  


       


            // Reducing divisor by a factor  


            // of 2 as 2 digits are dropped  


            divisor = divisor / 100;  


        }  


        return true;  


    }  


       


    // Function to return the largest  


    // palindromic prime present in the array  


    static int maxPalindromicPrime(int []arr, int n)  


    {  


        int maxElement = Arrays.stream(arr).max().getAsInt(); 


       


        // Create a boolean array "prime[0..n]" and  


        // initialize all entries it as true. A value  


        // in prime[i] will finally be false if i is  


        // Not a prime, else true.  


        boolean []prime = new boolean[maxElement + 1];  


        for (int i = 0; i < maxElement + 1 ; i++) 


            prime[i] = true ; 


   


        // 0 and 1 are not primes  


        prime[0] = prime[1] = false;  


        for (int p = 2; p * p <= maxElement; p++) {  


       


            // If prime[p] is not changed, then it is  


            // a prime  


            if (prime[p] == true) {  


       


                // Update all multiples of p  


                for (int i = p * 2; i <= maxElement; i += p)  


                    prime[i] = false;  


            }  


        }  


       


        int currentMax = -1;  


        for (int i = 0; i < n; i++)  


       


            // If arr[i] is prime as well as palindrome  


            if (prime[arr[i]] == true && isPal(arr[i]) == true)  


                currentMax = Math.max(currentMax, arr[i]);  


       


        return currentMax;  


    }  


       


    // Driver Program  


     public static void main(String []args) 


    {  


        int []arr = { 11, 5, 121, 7, 89 };  


        int n = arr.length ; 


        System.out.println(maxPalindromicPrime(arr, n)) ;  


    }  


       


} 


// This code is contributed by 29AjayKumar  



















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Python3














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# Python 3 implementation of the approach 


from math import sqrt 


  


# Function that returns true  


# if n is a palindrome 


def isPal(n): 


      


    # Find the appropriate divisor 


    # to extract the leading digit 


    divisor = 1


    while (n / divisor >= 10): 


        divisor *= 10


  


    while (n != 0): 


        leading = int(n / divisor) 


        trailing = n % 10


  


        # If first and last digit 


        # not same return false 


        if (leading != trailing): 


            return False


  


        # Removing the leading and trailing 


        # digit from number 


        n = int((n % divisor) / 10) 


  


        # Reducing divisor by a factor 


        # of 2 as 2 digits are dropped 


        divisor = int(divisor / 100) 


      


    return True


  


# Function to return the largest 


# palindromic prime present in the array 


def maxPalindromicPrime(arr, n): 


    maxElement = arr[0] 


    for i in range(len(arr)): 


        if (arr[i]>maxElement): 


            maxElement = arr[i] 


  


    # Create a boolean array "prime[0..n]" and 


    # initialize all entries it as true. A value 


    # in prime[i] will finally be false if i is 


    # Not a prime, else true. 


    prime = [True for i in range(maxElement + 1)] 


  


    # 0 and 1 are not primes 


    prime[0] = False


    prime[1] = False


    for p in range(2, int(sqrt(maxElement)) + 1, 1): 


          


        # If prime[p] is not changed, 


        # then it is a prime 


        if (prime[p] == True): 


              


            # Update all multiples of p 


            for i in range(p * 2,maxElement + 1, p): 


                prime[i] = False


      


    currentMax = -1


    for i in range(n): 


          


        # If arr[i] is prime as well as palindrome 


        if (prime[arr[i]] and isPal(arr[i])): 


            currentMax = max(currentMax, arr[i]) 


  


    return currentMax 


  


# Driver Code 


if __name__ == '__main__': 


    arr = [11, 5, 121, 7, 89] 


    n = len(arr) 


    print(maxPalindromicPrime(arr, n)) 


  


# This code is contributed by 


# Shashank_Shamra 



















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C#














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// C# implementation of the above approach  


  


using System ; 


using System.Linq ; 


  


public class GFG{ 


  


    // Function that returns true if n is a palindrome  


    static bool isPal(int n)  


    {  


        // Find the appropriate divisor  


        // to extract the leading digit  


        int divisor = 1;  


        while (n / divisor >= 10)  


            divisor *= 10;  


      


        while (n != 0) {  


            int leading = n / divisor;  


            int trailing = n % 10;  


      


            // If first and last digit  


            // not same return false  


            if (leading != trailing)  


                return false;  


      


            // Removing the leading and trailing  


            // digit from number  


            n = (n % divisor) / 10;  


      


            // Reducing divisor by a factor  


            // of 2 as 2 digits are dropped  


            divisor = divisor / 100;  


        }  


        return true;  


    }  


      


    // Function to return the largest  


    // palindromic prime present in the array  


    static int maxPalindromicPrime(int []arr, int n)  


    {  


        int maxElement = arr.Max() ; 


      


        // Create a boolean array "prime[0..n]" and  


        // initialize all entries it as true. A value  


        // in prime[i] will finally be false if i is  


        // Not a prime, else true.  


        bool []prime = new bool [maxElement + 1];  


        for (int i = 0; i < maxElement + 1 ; i++) 


            prime[i] = true ; 


  


        // 0 and 1 are not primes  


        prime[0] = prime[1] = false;  


        for (int p = 2; p * p <= maxElement; p++) {  


      


            // If prime[p] is not changed, then it is  


            // a prime  


            if (prime[p] == true) {  


      


                // Update all multiples of p  


                for (int i = p * 2; i <= maxElement; i += p)  


                    prime[i] = false;  


            }  


        }  


      


        int currentMax = -1;  


        for (int i = 0; i < n; i++)  


      


            // If arr[i] is prime as well as palindrome  


            if (prime[arr[i]] == true && isPal(arr[i]) == true)  


                currentMax = Math.Max(currentMax, arr[i]);  


      


        return currentMax;  


    }  


      


    // Driver Program  


     public static void Main() 


    {  


        int []arr = { 11, 5, 121, 7, 89 };  


        int n = arr.Length ; 


        Console.WriteLine(maxPalindromicPrime(arr, n)) ;  


    }  


      


    // This code is contributed by Ryuga 


} 



















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PHP





= 10)

		$divisor *= 10;


	while ($n != 0)

	{

		$leading = (int)($n / $divisor);

		$trailing = $n % 10;


		// If first and last digit

		// not same return false

		if ($leading != $trailing)

			return false;


		// Removing the leading and trailing

		// digit from number

		$n = (int)(($n % $divisor) / 10);


		// Reducing divisor by a factor

		// of 2 as 2 digits are dropped

		$divisor = (int)($divisor / 100);

	}

	return true;

}


// Function to return the largest

// palindromic prime present in the array

function maxPalindromicPrime($arr, $n)

{

	$maxElement = max($arr);


	// Create a boolean array “prime[0..n]” and

	// initialize all entries it as true. A value

	// in prime[i] will finally be false if i is

	// Not a prime, else true.

	$prime = array_fill(0, ($maxElement + 1), true);


	// 0 and 1 are not primes

	$prime[0] = $prime[1] = false;

	for ($p = 2; $p * $p <= $maxElement; $p++) 
	{

		// If prime[p] is not changed, 
		// then it is a prime
		if ($prime[$p] == true) 
		{

			// Update all multiples of p
			for ($i = $p * 2; 
			     $i <= $maxElement; $i += $p)
				$prime[$i] = false;
		}
	}

	$currentMax = -1;
	for ($i = 0; $i < $n; $i++)

		// If arr[i] is prime as well as palindrome
		if ($prime[$arr[$i]] && isPal($arr[$i]))
			$currentMax = max($currentMax, $arr[$i]);

	return $currentMax;
}

// Driver Code
$arr = array( 11, 5, 121, 7, 89 );
$n = count($arr);
echo maxPalindromicPrime($arr, $n);

// This code is contributed by mits
?>





Output:


11






          
          
          
          

            

    

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