Given an integer N, the task is to check if N is a Rare Number.
Rare Number is a number N which is non-palindromic and N+rev(N) and N-rev(N) are both perfect squares where rev(N) is the reverse of the number N. For Example rev(65) = 56
Examples:
Input: N = 65
Output: Yes
65 – 56 = 9 and 65 + 56 = 121 are both perfect squaresInput: N = 10
Output: No
Approach: The idea is to check if N is a palindromic number, then return false. And if it is non-palindromic then just check whether N + rev(N) and N – rev(N) are both perfect squares or not.
Below is the implementation of the above approach:
C++
// C++ implementation to check if // N is a Rare number#include<bits/stdc++.h>using namespace std;// Iterative function to // reverse digits of numint reverseDigits(int num) { int rev_num = 0; while(num > 0) { rev_num = rev_num*10 + num%10; num = num/10; } return rev_num; } // Function to check if N// is perfect squarebool isPerfectSquare(long double x) { // Find floating point value of // square root of x. long double sr = sqrt(x); // If square root is an integer return ((sr - floor(sr)) == 0); } // Function to check if N is an // Rare numberbool isRare(int N){ // Find reverse of N int reverseN = reverseDigits(N); // Number should be non-palindromic if(reverseN == N) return false; return isPerfectSquare(N + reverseN) && isPerfectSquare(N - reverseN);}// Driver Codeint main(){ int n = 65; if (isRare(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation to check if N// is a Rare numberclass GFG{ // Iterative function to // reverse digits of numstatic int reverseDigits(int num) { int rev_num = 0; while(num > 0) { rev_num = rev_num * 10 + num % 10; num = num / 10; } return rev_num; } // Function to check if N// is perfect squarestatic boolean isPerfectSquare(double x) { // Find floating point value of // square root of x. double sr = Math.sqrt(x); // If square root is an integer return ((sr - Math.floor(sr)) == 0); } // Function to check if N is an // Rare numberstatic boolean isRare(int N){ // Find reverse of N int reverseN = reverseDigits(N); // Number should be non-palindromic if(reverseN == N) return false; return isPerfectSquare(N + reverseN) && isPerfectSquare(N - reverseN);}// Driver code public static void main(String[] args) { int n = 65; if (isRare(n)) { System.out.println("Yes"); } else { System.out.println("No"); }} } // This code is contributed by shubham |
Python3
# Python3 implementation to check if # N is a Rare numberimport math# Iterative function to # reverse digits of numdef reverseDigits(num): rev_num = 0 while(num > 0): rev_num = rev_num * 10 + num % 10 num = num // 10 return rev_num# Function to check if N# is perfect squaredef isPerfectSquare(x): # Find floating point value of # square root of x. sr = math.sqrt(x) # If square root is an integer return ((sr - int(sr)) == 0)# Function to check if N is an # Rare numberdef isRare(N): # Find reverse of N reverseN = reverseDigits(N) # Number should be non-palindromic if(reverseN == N): return False return (isPerfectSquare(N + reverseN) and isPerfectSquare(N - reverseN))# Driver CodeN = 65if (isRare(N)): print("Yes")else: print("No")# This code is contributed by Vishal Maurya |
C#
// C# implementation to check if N// is a Rare numberusing System;class GFG{ // Iterative function to // reverse digits of numstatic int reverseDigits(int num) { int rev_num = 0; while(num > 0) { rev_num = rev_num * 10 + num % 10; num = num / 10; } return rev_num; } // Function to check if N// is perfect squarestatic bool isPerfectSquare(double x) { // Find floating point value of // square root of x. double sr = Math.Sqrt(x); // If square root is an integer return ((sr - Math.Floor(sr)) == 0); } // Function to check if N is an // Rare numberstatic bool isRare(int N){ // Find reverse of N int reverseN = reverseDigits(N); // Number should be non-palindromic if(reverseN == N) return false; return isPerfectSquare(N + reverseN) && isPerfectSquare(N - reverseN);} // Driver code public static void Main(String[] args) { int n = 65; if (isRare(n)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); }} }// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript implementation to check if N// is a Rare number // Iterative function to // reverse digits of num function reverseDigits( num) { let rev_num = 0; while (num > 0) { rev_num = rev_num * 10 + num % 10; num = parseInt(num / 10); } return rev_num; } // Function to check if N // is perfect square function isPerfectSquare( x) { // Find floating point value of // square root of x. let sr = Math.sqrt(x); // If square root is an integer return ((sr - Math.floor(sr)) == 0); } // Function to check if N is an // Rare number function isRare( N) { // Find reverse of N let reverseN = reverseDigits(N); // Number should be non-palindromic if (reverseN == N) return false; return isPerfectSquare(N + reverseN) && isPerfectSquare(N - reverseN); } // Driver code let n = 65; if (isRare(n)) { document.write("Yes"); } else { document.write("No"); }// This code is contributed by todaysgaurav </script> |
Output:
Yes
Time Complexity: O(N1/2)
References: OEIS
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