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++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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
}

PHP

<?php
// PHP implementation of the approach

// Function that returns true
// if n is a palindrome
function isPal($n)
{
    // Find the appropriate divisor
    // to extract the leading digit
    $divisor = 1;
    while ((int)($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
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
?>

Javascript

<script>
// Javascript implementation of the approach

// Function that returns true
// if n is a palindrome
function isPal(n) {
    // Find the appropriate divisor
    // to extract the leading digit
    let divisor = 1;
    while (Math.floor(n / divisor) >= 10)
        divisor *= 10;

    while (n != 0) {
        let leading = Math.floor(n / divisor);
        let 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 = Math.floor((n % divisor) / 10);

        // Reducing divisor by a factor
        // of 2 as 2 digits are dropped
        divisor = Math.floor(divisor / 100);
    }
    return true;
}

// Function to return the largest
// palindromic prime present in the array
function maxPalindromicPrime(arr, n) {
    let maxElement = arr.sort((a, b) => a - b).reverse()[0];

    // 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.
    let prime = new Array(maxElement + 1).fill(true);

    // 0 and 1 are not primes
    prime[0] = prime[1] = false;
    for (let 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 (let i = p * 2;
                i <= maxElement; i += p)
                prime[i] = false;
        }
    }

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

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

    return currentMax;
}

// Driver Code
let arr = [11, 5, 121, 7, 89];
let n = arr.length;
document.write(maxPalindromicPrime(arr, n));

// This code is contributed by gfgking
</script>

Output

11