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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 = 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(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     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 = False     prime = 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 ?>

Output:

11

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