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++
#include <bits/stdc++.h>
using namespace std;
bool isPal( int n)
{
int divisor = 1;
while (n / divisor >= 10)
divisor *= 10;
while (n != 0) {
int leading = n / divisor;
int trailing = n % 10;
if (leading != trailing)
return false ;
n = (n % divisor) / 10;
divisor = divisor / 100;
}
return true ;
}
int maxPalindromicPrime( int arr[], int n)
{
int maxElement = *max_element(arr, arr + n);
bool prime[maxElement + 1];
memset (prime, true , sizeof (prime));
prime[0] = prime[1] = false ;
for ( int p = 2; p * p <= maxElement; p++) {
if (prime[p] == true ) {
for ( int i = p * 2; i <= maxElement; i += p)
prime[i] = false ;
}
}
int currentMax = -1;
for ( int i = 0; i < n; i++)
if (prime[arr[i]] && isPal(arr[i]))
currentMax = max(currentMax, arr[i]);
return currentMax;
}
int main()
{
int arr[] = { 11, 5, 121, 7, 89 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << maxPalindromicPrime(arr, n);
return 0;
}
|
Java
import java.util.Arrays;
public class GFG{
static boolean isPal( int n)
{
int divisor = 1 ;
while (n / divisor >= 10 )
divisor *= 10 ;
while (n != 0 ) {
int leading = n / divisor;
int trailing = n % 10 ;
if (leading != trailing)
return false ;
n = (n % divisor) / 10 ;
divisor = divisor / 100 ;
}
return true ;
}
static int maxPalindromicPrime( int []arr, int n)
{
int maxElement = Arrays.stream(arr).max().getAsInt();
boolean []prime = new boolean [maxElement + 1 ];
for ( int i = 0 ; i < maxElement + 1 ; i++)
prime[i] = true ;
prime[ 0 ] = prime[ 1 ] = false ;
for ( int p = 2 ; p * p <= maxElement; p++) {
if (prime[p] == true ) {
for ( int i = p * 2 ; i <= maxElement; i += p)
prime[i] = false ;
}
}
int currentMax = - 1 ;
for ( int i = 0 ; i < n; i++)
if (prime[arr[i]] == true && isPal(arr[i]) == true )
currentMax = Math.max(currentMax, arr[i]);
return currentMax;
}
public static void main(String []args)
{
int []arr = { 11 , 5 , 121 , 7 , 89 };
int n = arr.length ;
System.out.println(maxPalindromicPrime(arr, n)) ;
}
}
|
Python3
from math import sqrt
def isPal(n):
divisor = 1
while (n / divisor > = 10 ):
divisor * = 10
while (n ! = 0 ):
leading = int (n / divisor)
trailing = n % 10
if (leading ! = trailing):
return False
n = int ((n % divisor) / 10 )
divisor = int (divisor / 100 )
return True
def maxPalindromicPrime(arr, n):
maxElement = arr[ 0 ]
for i in range ( len (arr)):
if (arr[i]>maxElement):
maxElement = arr[i]
prime = [ True for i in range (maxElement + 1 )]
prime[ 0 ] = False
prime[ 1 ] = False
for p in range ( 2 , int (sqrt(maxElement)) + 1 , 1 ):
if (prime[p] = = True ):
for i in range (p * 2 ,maxElement + 1 , p):
prime[i] = False
currentMax = - 1
for i in range (n):
if (prime[arr[i]] and isPal(arr[i])):
currentMax = max (currentMax, arr[i])
return currentMax
if __name__ = = '__main__' :
arr = [ 11 , 5 , 121 , 7 , 89 ]
n = len (arr)
print (maxPalindromicPrime(arr, n))
|
C#
using System ;
using System.Linq ;
public class GFG{
static bool isPal( int n)
{
int divisor = 1;
while (n / divisor >= 10)
divisor *= 10;
while (n != 0) {
int leading = n / divisor;
int trailing = n % 10;
if (leading != trailing)
return false ;
n = (n % divisor) / 10;
divisor = divisor / 100;
}
return true ;
}
static int maxPalindromicPrime( int []arr, int n)
{
int maxElement = arr.Max() ;
bool []prime = new bool [maxElement + 1];
for ( int i = 0; i < maxElement + 1 ; i++)
prime[i] = true ;
prime[0] = prime[1] = false ;
for ( int p = 2; p * p <= maxElement; p++) {
if (prime[p] == true ) {
for ( int i = p * 2; i <= maxElement; i += p)
prime[i] = false ;
}
}
int currentMax = -1;
for ( int i = 0; i < n; i++)
if (prime[arr[i]] == true && isPal(arr[i]) == true )
currentMax = Math.Max(currentMax, arr[i]);
return currentMax;
}
public static void Main()
{
int []arr = { 11, 5, 121, 7, 89 };
int n = arr.Length ;
Console.WriteLine(maxPalindromicPrime(arr, n)) ;
}
}
|
PHP
<?php
function isPal( $n )
{
$divisor = 1;
while ((int)( $n / $divisor ) >= 10)
$divisor *= 10;
while ( $n != 0)
{
$leading = (int)( $n / $divisor );
$trailing = $n % 10;
if ( $leading != $trailing )
return false;
$n = (int)(( $n % $divisor ) / 10);
$divisor = (int)( $divisor / 100);
}
return true;
}
function maxPalindromicPrime( $arr , $n )
{
$maxElement = max( $arr );
$prime = array_fill (0, ( $maxElement + 1), true);
$prime [0] = $prime [1] = false;
for ( $p = 2; $p * $p <= $maxElement ; $p ++)
{
if ( $prime [ $p ] == true)
{
for ( $i = $p * 2;
$i <= $maxElement ; $i += $p )
$prime [ $i ] = false;
}
}
$currentMax = -1;
for ( $i = 0; $i < $n ; $i ++)
if ( $prime [ $arr [ $i ]] && isPal( $arr [ $i ]))
$currentMax = max( $currentMax , $arr [ $i ]);
return $currentMax ;
}
$arr = array ( 11, 5, 121, 7, 89 );
$n = count ( $arr );
echo maxPalindromicPrime( $arr , $n );
?>
|
Javascript
<script>
function isPal(n) {
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 (leading != trailing)
return false ;
n = Math.floor((n % divisor) / 10);
divisor = Math.floor(divisor / 100);
}
return true ;
}
function maxPalindromicPrime(arr, n) {
let maxElement = arr.sort((a, b) => a - b).reverse()[0];
let prime = new Array(maxElement + 1).fill( true );
prime[0] = prime[1] = false ;
for (let p = 2; p * p <= maxElement; p++) {
if (prime[p] == true ) {
for (let i = p * 2;
i <= maxElement; i += p)
prime[i] = false ;
}
}
let currentMax = -1;
for (let i = 0; i < n; i++)
if (prime[arr[i]] && isPal(arr[i]))
currentMax = Math.max(currentMax, arr[i]);
return currentMax;
}
let arr = [11, 5, 121, 7, 89];
let n = arr.length;
document.write(maxPalindromicPrime(arr, n));
</script>
|
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[0], 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 been 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 <bits/stdc++.h>
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." <<endl;
} else {
cout << "The largest palindromic prime in the array is " << largest_pal_prime << "." <<endl;
}
return 0;
}
|
C
#include <stdio.h>
#include <stdbool.h>
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[0]);
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;
}
|
Java
import java.util.*;
public class Main {
public static boolean 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 ;
}
public static boolean is_palindrome( int n) {
int reversed = 0 ;
int original = n;
while (n > 0 ) {
reversed = reversed * 10 + n % 10 ;
n /= 10 ;
}
return (reversed == original);
}
public static void main(String[] args) {
List<Integer> arr = Arrays.asList( 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.get(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 ) {
System.out.println( "No palindromic prime found in the array." );
} else {
System.out.println( "The largest palindromic prime in the array is " + largest_pal_prime + "." );
}
}
}
|
Python3
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
def is_palindrome(n):
reversed = 0
original = n
while n > 0 :
reversed = reversed * 10 + n % 10
n / / = 10
return reversed = = original
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))
|
C#
using System;
using System.Collections.Generic;
public class PalindromePrimeFinder {
static bool IsPrime( int n)
{
if (n < 2) {
return false ;
}
for ( int i = 2; i * i <= n; i++) {
if (n % i == 0) {
return false ;
}
}
return true ;
}
static bool IsPalindrome( int n)
{
int reversed = 0;
int original = n;
while (n > 0) {
reversed = reversed * 10 + n % 10;
n /= 10;
}
return (reversed == original);
}
static void Main()
{
List< int > arr
= new List< int >{ 13, 101, 37, 313, 79,
181, 97, 131, 23, 199 };
int size = arr.Count;
int largest_pal_prime = -1;
for ( int i = 0; i < size; i++) {
int current_num = arr[i];
if (IsPrime(current_num)
&& IsPalindrome(current_num)) {
if (current_num > largest_pal_prime) {
largest_pal_prime = current_num;
}
}
}
if (largest_pal_prime == -1) {
Console.WriteLine(
"No palindromic prime found in the array." );
}
else {
Console.WriteLine(
"The largest palindromic prime in the array is "
+ largest_pal_prime + "." );
}
}
}
|
Javascript
function is_prime(n) {
if (n < 2) {
return false ;
}
for (let i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
return false ;
}
}
return true ;
}
function is_palindrome(n) {
let reversed = 0;
let original = n;
while (n > 0) {
reversed = reversed * 10 + n % 10;
n = Math.floor(n/10);
}
return reversed == original;
}
let arr = [13, 101, 37, 313, 79, 181, 97, 131, 23, 199];
let size = arr.length;
let largest_pal_prime = -1;
for (let i = 0; i < size; i++) {
let 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) {
console.log( "No palindromic prime found in the array." );
}
else {
console.log(`The largest palindromic prime in the array is ${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.
Last Updated :
31 Mar, 2023
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