Given an integer N, the task is to check if N is a Dihedral prime number or not. A Dihedral prime is a prime number that can be read as itself or as another prime number when read in a seven-segment display, regardless of different orientation and surface.
Examples:
Input: N = 108881
Output: YesInput: N = 789
Output: No
Approach: Pre-calculate prime number sieve for primality testing. The Sieves of Eratosthenes can be calculated in n*logn*logn time. Run a primality test for the number and its different orientations. If the number passes the primality tests, check if any digits belong to the exclusion set [3, 4, 6, 7, 9]. The return is true if the number passes both tests.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;bool isPrime[int(1e5) + 5];// Function to return the reverse// of a numberint reverse(int n){ int temp = n; int sum = 0; while (temp > 0) { int rem = temp % 10; sum = sum * 10 + rem; temp /= 10; } return sum;}// Function to generate mirror reflection// of a numberint mirror(int n){ int temp = n; int sum = 0; while (temp > 0) { int rem = temp % 10; if (rem == 2) rem = 5; else if (rem == 5) rem = 2; sum = sum * 10 + rem; temp /= 10; } return sum;}// Function to initialize prime number sievebool sieve(){ memset(isPrime, true, sizeof isPrime); isPrime[0] = isPrime[1] = false; for (int i = 2; i <= int(1e5); i++) { for (int j = 2; i * j <= int(1e5); j++) { isPrime[i * j] = false; } }}// Function that returns true if n is// Dihedral Prime numberbool isDihedralPrime(int n){ // Check if the orientations of n // is also prime if (!isPrime[n] || !isPrime[mirror(n)] || !isPrime[reverse(n)] || !isPrime[reverse(mirror(n))]) return false; int temp = n; while (temp > 0) { int rem = temp % 10; if (rem == 3 || rem == 4 || rem == 6 || rem == 7 || rem == 9) return false; temp /= 10; } return true;}// Driver codeint main(){ sieve(); int n = 18181; if (isDihedralPrime(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the above approach import java.util.*;class GFG { static boolean[] isPrime = new boolean[(int) (1e5) + 5]; // Function to return the reverse // of a number static int reverse(int n) { int temp = n; int sum = 0; while (temp > 0) { int rem = temp % 10; sum = sum * 10 + rem; temp /= 10; } return sum; } // Function to generate mirror reflection // of a number static int mirror(int n) { int temp = n; int sum = 0; while (temp > 0) { int rem = temp % 10; if (rem == 2) { rem = 5; } else if (rem == 5) { rem = 2; } sum = sum * 10 + rem; temp /= 10; } return sum; } // Function to initialize // prime number sieve static void sieve() { Arrays.fill(isPrime, true); isPrime[0] = isPrime[1] = false; for (int i = 2; i <= (int) 1e5; i++) { for (int j = 2; i * j <= (int) 1e5; j++) { isPrime[i * j] = false; } } } // Function that returns true if n is // Dihedral Prime number static boolean isDihedralPrime(int n) { // Check if the orientations of n // is also prime if (!isPrime[n] || !isPrime[mirror(n)] || !isPrime[reverse(n)] || !isPrime[reverse(mirror(n))]) { return false; } int temp = n; while (temp > 0) { int rem = temp % 10; if (rem == 3 || rem == 4 || rem == 6 || rem == 7 || rem == 9) { return false; } temp /= 10; } return true; } // Driver code public static void main(String[] args) { sieve(); int n = 18181; if (isDihedralPrime(n)) { System.out.println("Yes"); } else { System.out.println("No"); } }}// This code is contributed by Rajput-Ji |
Python3
# Python implementation of the above approachisPrime = (int(1e5)+5)*[True]# Function to return the reverse# of a numberdef reverse(n): temp = n sum = 0 while temp>0: rem = temp % 10 sum = sum * 10 + rem temp//= 10 return sum# Function to generate mirror reflection# of a numberdef mirror(n): temp = n sum = 0 while temp>0: rem = temp % 10 if rem == 2: rem = 5 elif rem == 5: rem = 2 sum = sum * 10 + rem temp//= 10 return sum# Function to initialize prime number sievedef sieve(): isPrime[0] = isPrime[1] = False for i in range(2, int(1e5)+1): j = 2 while i * j<= int(1e5): isPrime[i * j] = False j+= 1# Function that returns true if n is# Dihedral Prime numberdef isDihedralPrime(n): # Check if the orientations of n is also prime if (not isPrime[n]) or (not isPrime[mirror(n)]) \ or (not isPrime[reverse(n)]) \ or (not isPrime[reverse(mirror(n))]): return False temp = n while temp>0: rem = temp % 10; if rem == 3 or rem == 4 or \ rem == 6 or rem == 7 or rem == 9: return False temp//= 10 return True# Driver codeif __name__ == '__main__': sieve() n = 18181 if isDihedralPrime(n): print("Yes") else : print("No") |
C#
// C# implementation of the above approachusing System; class GFG { static Boolean[] isPrime = new Boolean[(int) (1e5) + 5]; // Function to return the reverse // of a number static int reverse(int n) { int temp = n; int sum = 0; while (temp > 0) { int rem = temp % 10; sum = sum * 10 + rem; temp /= 10; } return sum; } // Function to generate mirror reflection // of a number static int mirror(int n) { int temp = n; int sum = 0; while (temp > 0) { int rem = temp % 10; if (rem == 2) { rem = 5; } else if (rem == 5) { rem = 2; } sum = sum * 10 + rem; temp /= 10; } return sum; } // Function to initialize // prime number sieve static void sieve() { for(int k = 0; k < isPrime.Length; k++) isPrime[k] = true; isPrime[0] = isPrime[1] = false; for (int i = 2; i <= (int) 1e5; i++) { for (int j = 2; i * j <= (int) 1e5; j++) { isPrime[i * j] = false; } } } // Function that returns true if n is // Dihedral Prime number static Boolean isDihedralPrime(int n) { // Check if the orientations of n // is also prime if (!isPrime[n] || !isPrime[mirror(n)] || !isPrime[reverse(n)] || !isPrime[reverse(mirror(n))]) { return false; } int temp = n; while (temp > 0) { int rem = temp % 10; if (rem == 3 || rem == 4 || rem == 6 || rem == 7 || rem == 9) { return false; } temp /= 10; } return true; } // Driver code public static void Main(String[] args) { sieve(); int n = 18181; if (isDihedralPrime(n)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }}// This code is contributed by Rajput-Ji |
PHP
<?php// PHP implementation of the above approach $isPrime = array_fill(0, 25000 , true); // Function to return the reverse // of a number function reverse($n) { $temp = $n; $sum = 0; while ($temp > 0) { $rem = $temp % 10; $sum = $sum * 10 + $rem; $temp = floor($temp / 10); } return $sum; } // Function to generate mirror reflection // of a number function mirror($n) { $temp = $n; $sum = 0; while ($temp > 0) { $rem = $temp % 10; if ($rem == 2) $rem = 5; else if ($rem == 5) $rem = 2; $sum = $sum * 10 + $rem; $temp = floor($temp / 10); } return $sum; } // Function to initialize prime number sieve function sieve() { $GLOBALS['isPrime'][0] = $GLOBALS['isPrime'][1] = false; for ($i = 2; $i <= floor(1e4); $i++) { for ($j = 2; $i * $j <= floor(1e4); $j++) { $GLOBALS['isPrime'][$i * $j] = false; } } } // Function that returns true if n is // Dihedral Prime number function isDihedralPrime($n) { // Check if the orientations of n // is also prime if (!$GLOBALS['isPrime'][$n] || !$GLOBALS['isPrime'][mirror($n)] || !$GLOBALS['isPrime'][reverse($n)] || !$GLOBALS['isPrime'][reverse(mirror($n))]) return false; $temp = $n; while ($temp > 0) { $rem = $temp % 10; if ($rem == 3 || $rem == 4 || $rem == 6 || $rem == 7 || $rem == 9) return false; $temp = floor($temp / 10); } return true; } // Driver code sieve(); $n = 18181; if (isDihedralPrime($n)) echo "Yes"; else echo "No"; // This code is contributed by Ryuga ?> |
Javascript
<script>// Javascript implementation of the above approachlet isPrime = new Array(25000).fill(true);// Function to return the reverse// of a numberfunction reverse(n){ let temp = n; let sum = 0; while (temp > 0) { rem = temp % 10; sum = sum * 10 + rem; temp = Math.floor(temp / 10); } return sum;}// Function to generate mirror reflection// of a numberfunction mirror(n){ let temp = n; let sum = 0; while (temp > 0) { let rem = temp % 10; if (rem == 2) rem = 5; else if (rem == 5) rem = 2; sum = sum * 10 + rem; temp = Math.floor(temp / 10); } return sum;}// Function to initialize prime number sievefunction sieve(){ isPrime[0] = isPrime[1] = false; for (let i = 2; i <= Math.floor(1e4); i++) { for (let j = 2; i * j <= Math.floor(1e4); j++) { isPrime[i * j] = false; } }}// Function that returns true if n is// Dihedral Prime numberfunction isDihedralPrime(n){ // Check if the orientations of n // is also prime if (!isPrime[n] || !isPrime[mirror(n)] || !isPrime[reverse(n)] || !isPrime[reverse(mirror(n))]) return false; let temp = n; while (temp > 0) { rem = temp % 10; if (rem == 3 || rem == 4 || rem == 6 || rem == 7 || rem == 9) return false; temp = Math.floor(temp / 10); } return true;}// Driver codesieve();let n = 18181;if (isDihedralPrime(n)) document.write( "Yes");else document.write( "No");// This code is contributed by Saurabh Jaiswal</script> |
Output:
Yes
Time Complexity: O(n * log(log n))
Auxiliary Space: O(105)
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