Rare Numbers

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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 squares

Input: 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 num
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 square
bool 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 number
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
int main()
{
    int n = 65;
    if (isRare(n))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java implementation to check if N
// is a Rare number
class GFG{ 
 
// Iterative function to 
// reverse digits of num
static 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 square
static 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 number
static 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 number
import math
 
# Iterative function to 
# reverse digits of num
def 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 square
def 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 number
def 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 Code
N = 65
if (isRare(N)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Vishal Maurya

C#

// C# implementation to check if N
// is a Rare number
using System;
class GFG{ 
  
// Iterative function to 
// reverse digits of num
static 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 square
static 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 number
static 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(N^1/2)
References: OEIS

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