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# Largest palindromic prime in an array

• Difficulty Level : Easy
• Last Updated : 20 Mar, 2023

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:

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 ``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 = prime = ``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);``    ``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 = prime = ``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

 `= 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`` = ``\$prime`` = 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

 ``

Output

`11`

#### Another Approach:

Define two helper functions: is_prime() and is_palindrome(). The is_prime() function checks if a given number is prime or not using a basic primality test, and returns a boolean value. The is_palindrome() function checks if a given number is a palindrome or not, and returns a boolean value.

Define an array of integers arr containing some values for the purpose of example. Calculate the size of the array using sizeof() and arr, and store it in the variable size.

Initialize the variable largest_pal_prime to -1. This is the default value to be used in case there are no palindromic primes found in the array.

Use a for loop to iterate over each element in the array. For each element, check if it is a palindrome and a prime using the helper functions. If it is both a palindrome and a prime, and its value is greater than the current largest_pal_prime, set largest_pal_prime to the value of the current element.

After the loop has completed, check if largest_pal_prime is still equal to -1. If it is, print a message indicating that no palindromic primes were found in the array. Otherwise, print the value of largest_pal_prime.

## C++

 `#include ``using` `namespace` `std;` `bool` `is_prime(``int` `n) {``    ``if` `(n < 2) {``        ``return` `false``;``    ``}``    ``for` `(``int` `i = 2; i * i <= n; i++) {``        ``if` `(n % i == 0) {``            ``return` `false``;``        ``}``    ``}``    ``return` `true``;``}` `bool` `is_palindrome(``int` `n) {``    ``int` `reversed = 0;``    ``int` `original = n;` `    ``while` `(n > 0) {``        ``reversed = reversed * 10 + n % 10;``        ``n /= 10;``    ``}` `    ``return` `(reversed == original);``}` `int` `main() {``    ``vector<``int``> arr = {13, 101, 37, 313, 79, 181, 97, 131, 23, 199};``    ``int` `size = arr.size();``    ``int` `largest_pal_prime = -1;` `    ``for` `(``int` `i = 0; i < size; i++) {``        ``int` `current_num = arr[i];``        ``if` `(is_prime(current_num) && is_palindrome(current_num)) {``            ``if` `(current_num > largest_pal_prime) {``                ``largest_pal_prime = current_num;``            ``}``        ``}``    ``}` `    ``if` `(largest_pal_prime == -1) {``        ``cout << ``"No palindromic prime found in the array."` `<

## C

 `#include ``#include ` `bool` `is_prime(``int` `n) {``    ``if` `(n < 2) {``        ``return` `false``;``    ``}``    ``for` `(``int` `i = 2; i * i <= n; i++) {``        ``if` `(n % i == 0) {``            ``return` `false``;``        ``}``    ``}``    ``return` `true``;``}` `bool` `is_palindrome(``int` `n) {``    ``int` `reversed = 0;``    ``int` `original = n;` `    ``while` `(n > 0) {``        ``reversed = reversed * 10 + n % 10;``        ``n /= 10;``    ``}` `    ``return` `(reversed == original);``}` `int` `main() {``    ``int` `arr[] = {13, 101, 37, 313, 79, 181, 97, 131, 23, 199};``    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `largest_pal_prime = -1;` `    ``for` `(``int` `i = 0; i < size; i++) {``        ``int` `current_num = arr[i];``        ``if` `(is_prime(current_num) && is_palindrome(current_num)) {``            ``if` `(current_num > largest_pal_prime) {``                ``largest_pal_prime = current_num;``            ``}``        ``}``    ``}` `    ``if` `(largest_pal_prime == -1) {``        ``printf``(``"No palindromic prime found in the array.\n"``);``    ``} ``else` `{``        ``printf``(``"The largest palindromic prime in the array is %d.\n"``, largest_pal_prime);``    ``}` `    ``return` `0;``}`

## Python3

 `# Python3 Program to implement``# the above approach` `# function to check if a number is prime``def` `is_prime(n):``    ``if` `n < ``2``:``        ``return` `False``    ``for` `i ``in` `range``(``2``, ``int``(n ``*``*` `0.5``) ``+` `1``):``        ``if` `n ``%` `i ``=``=` `0``:``            ``return` `False``    ``return` `True` `# function to check if a number is palindrome``def` `is_palindrome(n):``    ``reversed` `=` `0``    ``original ``=` `n``    ``while` `n > ``0``:``        ``reversed` `=` `reversed` `*` `10` `+` `n ``%` `10``        ``n ``/``/``=` `10``    ` `    ``return` `reversed` `=``=` `original` `# Driver code``arr ``=` `[``13``, ``101``, ``37``, ``313``, ``79``, ``181``, ``97``, ``131``, ``23``, ``199``]``size ``=` `len``(arr)``largest_pal_prime ``=` `-``1` `for` `i ``in` `range``(size):``    ``current_num ``=` `arr[i]``    ``if` `is_prime(current_num) ``and` `is_palindrome(current_num):``        ``if` `current_num > largest_pal_prime:``            ``largest_pal_prime ``=` `current_num` `if` `largest_pal_prime ``=``=` `-``1``:``    ``print``(``"No palindromic prime found in the array."``)``else``:``    ``print``(``"The largest palindromic prime in the array is {}."``.``format``(largest_pal_prime))`

Output

`The largest palindromic prime in the array is 313.`

The time complexity of this program is O(n*sqrt(n)), where n is the size of the array. This is because the is_prime() function performs a loop from 2 to the square root of the input number, and this loop is executed for each element in the array.

The space complexity is O(1), as the program only uses a fixed amount of memory to store the variables and the array.

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