# Check if a number is an Unusual Number or not

Given a positive integer N. The task is to check if N is an unusual number or not. Print ‘YES’ if M is an unusual number else print ‘NO’.

Unusual number : In Mathematics, an unusual number is a natural number whose greatest prime factor is strictly greater than square root of n.

The first few unusual numbers are –

2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 20, 21, 22, 23, 26, 28, 29, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 44, 46, 47, 51

Examples:

Input : N = 14
Output : YES
Explanation : 7 is largest prime factor of 14
and 7 is strictly greater than square root of 14

Input : N = 16
Output : NO
Explanation : 2 is largest prime factor of 16
and 2 is less than square root of 16 ( i.e 4 ).


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach :

1. Find the largest prime factor of the given number N. To find the largest prime factor of N refer this .
2. Check if the largest prime factor of N is strictly grater than square root of N.
3. If ‘YES’ then N is an Unusual number otherwise Not.

Below is the implementation of the above approach:

## C++

 // C++ Program to check Unusual number   #include    using namespace std;        // Utility function to find largest   // prime factor of a number   int largestPrimeFactor(int n)   {       // Initialize the maximum prime factor       // variable with the lowest one       int max = -1;            // Print the number of 2s that divide n       while (n % 2 == 0) {           max = 2;           n >>= 1; // equivalent to n /= 2       }            // n must be odd at this point, thus skip       // the even numbers and iterate only for       // odd integers       for (int i = 3; i <= sqrt(n); i += 2) {           while (n % i == 0) {               max = i;               n = n / i;           }       }            // This condition is to handle the case       // when n is a prime number greater than 2       if (n > 2)           max = n;            return max;   }        // Function to check Unusual number   bool checkUnusual(int n)   {       // Get the largest Prime Factor       // of the number       int factor = largestPrimeFactor(n);            // Check if largest prime factor       // is greater than sqrt(n)       if (factor > sqrt(n)) {           return true;       }       else {           return false;       }   }        // Driver Code   int main()   {       int n = 14;            if (checkUnusual(n)) {           cout << "YES"              << "\n";       }       else {           cout << "NO"              << "\n";       }            return 0;   }

## Java

 // Java Program to check Unusual number      class GFG {          // Utility function to find largest      // prime factor of a number      static int largestPrimeFactor(int n)      {          // Initialize the maximum prime factor          // variable with the lowest one          int max = -1;              // Print the number of 2s that divide n          while (n % 2 == 0) {              max = 2;              n >>= 1; // equivalent to n /= 2          }              // n must be odd at this point, thus skip          // the even numbers and iterate only for          // odd integers          for (int i = 3; i <= Math.sqrt(n); i += 2) {              while (n % i == 0) {                  max = i;                  n = n / i;              }          }              // This condition is to handle the case          // when n is a prime number greater than 2          if (n > 2)              max = n;              return max;      }          // Function to check Unusual number      static boolean checkUnusual(int n)      {          // Get the largest Prime Factor          // of the number          int factor = largestPrimeFactor(n);              // Check if largest prime factor          // is greater than sqrt(n)          if (factor > Math.sqrt(n)) {              return true;          }          else {              return false;          }      }          // Driver Code      public static void main(String[] args)      {          int n = 14;              if (checkUnusual(n)) {              System.out.println("YES");          }          else {              System.out.println("NO");          }      }  }

## Python3

 # Python Program to check Unusual number   from math import sqrt       # Utility function to find largest   # prime factor of a number   def largestPrimeFactor(n):         # Initialize the maximum prime factor       # variable with the lowest one       max = -1          # Print the number of 2s that divide n       while n % 2 == 0:          max = 2;           n >>= 1; # equivalent to n /= 2            # n must be odd at this point, thus skip       # the even numbers and iterate only for       # odd integers       for i in range(3,int(sqrt(n))+1,2):           while n % i == 0:              max = i;               n = n / i;            # This condition is to handle the case       # when n is a prime number greater than 2       if n > 2:          max = n           return max       # Function to check Unusual number   def checkUnusual(n):      # Get the largest Prime Factor       # of the number       factor = largestPrimeFactor(n)            # Check if largest prime factor       # is greater than sqrt(n)       if factor > sqrt(n):           return True     else :          return False      # Driver Code   if __name__ == '__main__':             n = 14          if checkUnusual(n):           print("YES")      else:          print("NO")     # This code is contributed   # by Harshit Saini

## C#

 // C# Program to check Unusual number      using System;  class GFG {          // Utility function to find largest      // prime factor of a number      static int largestPrimeFactor(int n)      {          // Initialize the maximum prime factor          // variable with the lowest one          int max = -1;              // Print the number of 2s that divide n          while (n % 2 == 0) {              max = 2;              n >>= 1; // equivalent to n /= 2          }              // n must be odd at this point, thus skip          // the even numbers and iterate only for          // odd integers          for (int i = 3; i <= Math.Sqrt(n); i += 2) {              while (n % i == 0) {                  max = i;                  n = n / i;              }          }              // This condition is to handle the case          // when n is a prime number greater than 2          if (n > 2)              max = n;              return max;      }          // Function to check Unusual number      static bool checkUnusual(int n)      {          // Get the largest Prime Factor          // of the number          int factor = largestPrimeFactor(n);              // Check if largest prime factor          // is greater than sqrt(n)          if (factor > Math.Sqrt(n)) {              return true;          }          else {              return false;          }      }          // Driver Code      public static void Main()      {          int n = 14;              if (checkUnusual(n)) {              Console.WriteLine("YES");          }          else {              Console.WriteLine("NO");          }      }  }

## PHP

 >= 1; // equivalent to n /= 2       }            // n must be odd at this point, thus skip       // the even numbers and iterate only for       // odd integers       for ($i = 3; $i <= sqrt($n); $i += 2) {           while ($n % $i == 0) {               $max = $i;               $n = $n / $i;   }   }     // This condition is to handle the case   // when n is a prime number greater than 2   if ($n > 2)           $max = $n;            return $max;  }    // Function to check Unusual number  function checkUnusual($n)   {       // Get the largest Prime Factor       // of the number       $factor = largestPrimeFactor($n);            // Check if largest prime factor      // is greater than sqrt(n)       if ($factor > sqrt($n)) {           return true;       }       else {           return false;       }   }        // Driver Code    $n = 14;    if (checkUnusual($n)) {       echo "YES"."\n";   }   else {       echo "NO"."\n";   }   // This code is contributed   // by Harshit Saini   ?>

Output:

YES


Time complexity:
Auxiliary space:

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