Given two integer N1 and N2, the task is to check if both the pairs are ormiston prime.
Ormiston prime is those numbers that are prime and have the same digits in a different order.
The first few Ormiston prime pairs are:
(1913, 1931), (18379, 18397), (19013, 19031), (25013, 25031) ……etc.
Examples:
Input: N1 = 1913, N2 = 1931
Output: YESInput: n1 = 5, n2 = 7
Output: NO
Approach: The idea is to check that both and N1 and N2 are primes and anagram or not of each other. If the numbers are prime and anagram of each other then the number is Ormiston prime pairs.
Below is the implementation of the above approach:
C++
// C++ implementation to // check Ormiston prime#include <iostream>using namespace std;// Function to check if the// number is a prime or notbool isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}const int TEN = 10;// Function to update the frequency array// such that freq[i] stores the// frequency of digit i in nvoid updateFreq(int n, int freq[]){ // While there are digits // left to process while (n) { int digit = n % TEN; // Update the frequency of // the current digit freq[digit]++; // Remove the last digit n /= TEN; }}// Function that returns true if a and b// are anagrams of each otherbool areAnagrams(int a, int b){ // To store the frequencies of // the digits in a and b int freqA[TEN] = { 0 }; int freqB[TEN] = { 0 }; // Update the frequency of // the digits in a updateFreq(a, freqA); // Update the frequency of // the digits in b updateFreq(b, freqB); // Match the frequencies of // the common digits for (int i = 0; i < TEN; i++) { // If frequency differs for any digit // then the numbers are not // anagrams of each other if (freqA[i] != freqB[i]) return false; } return true;}// Returns true if n1 and // n2 are Ormiston primesbool OrmistonPrime(int n1, int n2){ return (isPrime(n1) && isPrime(n2) && areAnagrams(n1, n2));}// Driver codeint main(){ int n1 = 1913, n2 = 1931; if (OrmistonPrime(n1, n2)) cout << "YES" << endl; else cout << "NO" << endl; return 0;} |
Java
// Java implementation to // check Ormiston primeimport java.util.*;class GFG{ // Function to check if the// number is a prime or notstatic boolean isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;} static int TEN = 10; // Function to update the frequency array// such that freq[i] stores the// frequency of digit i in nstatic void updateFreq(int n, int freq[]){ // While there are digits // left to process while (n > 0) { int digit = n % TEN; // Update the frequency of // the current digit freq[digit]++; // Remove the last digit n /= TEN; }} // Function that returns true if a and b// are anagrams of each otherstatic boolean areAnagrams(int a, int b){ // To store the frequencies of // the digits in a and b int freqA[] = new int[TEN]; int freqB[] = new int[TEN]; // Update the frequency of // the digits in a updateFreq(a, freqA); // Update the frequency of // the digits in b updateFreq(b, freqB); // Match the frequencies of // the common digits for (int i = 0; i < TEN; i++) { // If frequency differs for any digit // then the numbers are not // anagrams of each other if (freqA[i] != freqB[i]) return false; } return true;} // Returns true if n1 and // n2 are Ormiston primesstatic boolean OrmistonPrime(int n1, int n2){ return (isPrime(n1) && isPrime(n2) && areAnagrams(n1, n2));} // Driver codepublic static void main(String[] args){ int n1 = 1913, n2 = 1931; if (OrmistonPrime(n1, n2)) System.out.print("YES" +"\n"); else System.out.print("NO" +"\n");}}// This code is contributed by sapnasingh4991 |
Python3
# Python3 implementation to # check Ormiston prime # Function to check if the # number is a prime or not def isPrime(n): # Corner cases if (n <= 1): return False if (n <= 3): return True # This is checked so that we can skip # middle five numbers in below loop if (n % 2 == 0 or n % 3 == 0): return False i = 5 while(i * i <= n): if (n % i == 0 or n % (i + 2) == 0): return False i = i + 6 return TrueTEN = 10# Function to update the frequency array # such that freq[i] stores the # frequency of digit i in n def updateFreq(n, freq): # While there are digits # left to process while (n): digit = n % TEN # Update the frequency of # the current digit freq[digit] += 1 # Remove the last digit n = n // TEN# Function that returns true if a and b # are anagrams of each other def areAnagrams(a, b): # To store the frequencies of # the digits in a and b freqA = [0] * TEN freqB = [0] * TEN # Update the frequency of # the digits in a updateFreq(a, freqA) # Update the frequency of # the digits in b updateFreq(b, freqB) # Match the frequencies of # the common digits for i in range(TEN): # If frequency differs for any digit # then the numbers are not # anagrams of each other if (freqA[i] != freqB[i]): return False return True# Returns true if n1 and # n2 are Ormiston primes def OrmistonPrime(n1, n2): return (isPrime(n1) and isPrime(n2) and areAnagrams(n1, n2)) # Driver coden1, n2 = 1913, 1931if (OrmistonPrime(n1, n2)): print("YES") else: print("NO") # This code is contributed by divyeshrabadiya07 |
C#
// C# implementation to // check Ormiston primeusing System;class GFG{// Function to check if the// number is a prime or notstatic bool isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for(int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}static int TEN = 10;// Function to update the frequency array// such that freq[i] stores the// frequency of digit i in nstatic void updateFreq(int n, int []freq){ // While there are digits // left to process while (n > 0) { int digit = n % TEN; // Update the frequency of // the current digit freq[digit]++; // Remove the last digit n /= TEN; }}// Function that returns true if a and b// are anagrams of each otherstatic bool areAnagrams(int a, int b){ // To store the frequencies of // the digits in a and b int []freqA = new int[TEN]; int []freqB = new int[TEN]; // Update the frequency of // the digits in a updateFreq(a, freqA); // Update the frequency of // the digits in b updateFreq(b, freqB); // Match the frequencies of // the common digits for(int i = 0; i < TEN; i++) { // If frequency differs for any // digit then the numbers are not // anagrams of each other if (freqA[i] != freqB[i]) return false; } return true;}// Returns true if n1 and // n2 are Ormiston primesstatic bool OrmistonPrime(int n1, int n2){ return (isPrime(n1) && isPrime(n2) && areAnagrams(n1, n2));}// Driver codepublic static void Main(String[] args){ int n1 = 1913, n2 = 1931; if (OrmistonPrime(n1, n2)) Console.Write("YES" + "\n"); else Console.Write("NO" + "\n");}}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript implementation to // check Ormiston prime // Function to check if the // number is a prime or not function isPrime( n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (let i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; } const TEN = 10; // Function to update the frequency array // such that freq[i] stores the // frequency of digit i in n function updateFreq( n, freq) { // While there are digits // left to process while (n > 0) { let digit = n % TEN; // Update the frequency of // the current digit freq[digit]++; // Remove the last digit n = parseInt(n/TEN); } } // Function that returns true if a and b // are anagrams of each other function areAnagrams( a ,b) { // To store the frequencies of // the digits in a and b let freqA = Array(TEN).fill(0); let freqB = Array(TEN).fill(0); // Update the frequency of // the digits in a updateFreq(a, freqA); // Update the frequency of // the digits in b updateFreq(b, freqB); // Match the frequencies of // the common digits for ( i = 0; i < TEN; i++) { // If frequency differs for any digit // then the numbers are not // anagrams of each other if (freqA[i] != freqB[i]) return false; } return true; } // Returns true if n1 and // n2 are Ormiston primes function OrmistonPrime( n1, n2) { return (isPrime(n1) && isPrime(n2) && areAnagrams(n1, n2)); } // Driver code let n1 = 1913, n2 = 1931; if (OrmistonPrime(n1, n2)) document.write("YES" + "\n"); else document.write("NO" + "\n");// This code is contributed by todaysgaurav </script> |
Output:
YES
Time Complexity: O(N)
Reference: OEIS
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