Given an integer N, the task is to check if N is a prime number in Flipped Down, Mirror Flipped and Mirror Flipped Down forms of the given number.
Examples:
Input: N = 120121
Output: Yes
Explanation:
Flipped forms of the number:
Flipped Upside Down – 151051
Mirror Flipped – 121021
Mirror Upside Down – 150151
Since 1510151 and 121021 are both prime numbers, the flipped numbers are prime.Input: N = 12
Output: No
Explanation:
Flipped forms of the number:
Flipped Upside Down – 15
Mirror Flipped – 21
Mirror Upside Down – 51
All the flipped numbers are not prime.
Approach: Follow the steps below to solve the problem:
- Since the number N has to be prime in each of the Flipped upside down, Mirror Flipped and Mirror Upside Down forms, the only possible digits it should contain are {0, 1, 2, 5, 8}.
- Therefore, the problem reduces to check if the number is prime or not and if it is made up of the digits 0, 1, 2, 5, and 8 only.
- If found to be true, print “Yes“.
- Otherwise, print “No“.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to check if N// contains digits// 0, 1, 2, 5, 8 onlybool checkDigits(int n){ // Extract digits of N do { int r = n % 10; // Return false if any of // these digits are present if (r == 3 || r == 4 || r == 6 || r == 7 || r == 9) return false; n /= 10; } while (n != 0); return true;}// Function to check if// N is prime or notbool isPrime(int n){ if (n <= 1) return false; // Check for all factors for (int i = 2; i * i <= n; i++) { if (n % i == 0) return false; } return true;}// Function to check if n is prime// in all the desired formsint isAllPrime(int n){ return isPrime(n) && checkDigits(n);}// Driver Codeint main(){ int N = 101; if (isAllPrime(N)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to check if N// contains digits// 0, 1, 2, 5, 8 onlystatic boolean checkDigits(int n){ // Extract digits of N do { int r = n % 10; // Return false if any of // these digits are present if (r == 3 || r == 4 || r == 6 || r == 7 || r == 9) return false; n /= 10; } while (n != 0); return true;}// Function to check if// N is prime or notstatic boolean isPrime(int n){ if (n <= 1) return false; // Check for all factors for (int i = 2; i * i <= n; i++) { if (n % i == 0) return false; } return true;}// Function to check if n is prime// in all the desired formsstatic boolean isAllPrime(int n){ return isPrime(n) && checkDigits(n);}// Driver Codepublic static void main(String[] args){ int N = 101; if (isAllPrime(N)) System.out.print("Yes"); else System.out.print("No");}}// This code is contributed by gauravrajput1 |
Python3
# Python3 program to implement# the above approach# Function to check if N# contains digits# 0, 1, 2, 5, 8 onlydef checkDigits(n): # Extract digits of N while True: r = n % 10 # Return false if any of # these digits are present if (r == 3 or r == 4 or r == 6 or r == 7 or r == 9): return False n //= 10 if n == 0: break return True# Function to check if# N is prime or notdef isPrime(n): if (n <= 1): return False # Check for all factors for i in range(2, n + 1): if i * i > n: break if (n % i == 0): return False return True# Function to check if n is prime# in all the desired formsdef isAllPrime(n): return isPrime(n) and checkDigits(n)# Driver Codeif __name__ == '__main__': N = 101 if (isAllPrime(N)): print("Yes") else: print("No")# This code is contributed by mohit kumar 29 |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to check if N// contains digits// 0, 1, 2, 5, 8 onlystatic bool checkDigits(int n){ // Extract digits of N do { int r = n % 10; // Return false if any of // these digits are present if (r == 3 || r == 4 || r == 6 || r == 7 || r == 9) return false; n /= 10; } while (n != 0); return true;}// Function to check if// N is prime or notstatic bool isPrime(int n){ if (n <= 1) return false; // Check for all factors for (int i = 2; i * i <= n; i++) { if (n % i == 0) return false; } return true;}// Function to check if n is prime// in all the desired formsstatic bool isAllPrime(int n){ return isPrime(n) && checkDigits(n);}// Driver Codepublic static void Main(String[] args){ int N = 101; if (isAllPrime(N)) Console.Write("Yes"); else Console.Write("No");}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program to implement// the above approach// Function to check if N// contains digits// 0, 1, 2, 5, 8 onlyfunction checkDigits(n){ // Extract digits of N do { var r = n % 10; // Return false if any of // these digits are present if (r == 3 || r == 4 || r == 6 || r == 7 || r == 9) return false; n = parseInt(n/10); } while (n != 0); return true;}// Function to check if// N is prime or notfunction isPrime(n){ if (n <= 1) return false; // Check for all factors for (var i = 2; i * i <= n; i++) { if (n % i == 0) return false; } return true;}// Function to check if n is prime// in all the desired formsfunction isAllPrime(n){ return isPrime(n) && checkDigits(n);}// Driver Codevar N = 101;if (isAllPrime(N)) document.write("Yes");else document.write("No");</script> |
Output:
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!