number_of_divisors.java

/*
Given a Number. Find it's Number of Divisors.
The Number can be big such as 10^6 to 10^9
So, If we do this in O(N) we will get Time Limit Exceeded [TLE]
Better Approach is run a loop till sqrt(Number)
Example square root of 10^9 is 31622 (approx), so iteration optimized a lot.
*/

import java.util.Scanner;
import java.lang.*;
import java.math.*;

public class NumberofDivisors
{
   // this get_number_of_divisors will count the number of divisors of the numbe
   static long get_number_of_divisors(long number)
   {
      long number_of_divisors = 0, i = 0;
      int limit = (int) Math.sqrt(number);
      for(i = 1; i < limit; i++)
      {
         if(number % i == 0)
         {
            /* Here i is a divisor of that number
            ( Number / i ) is also a divisor
            */
            number_of_divisors += 2;
         }
      }
	

    if(number % limit == 0)
    {
        if(limit * limit == number)
        {
            //means perfect square number
           number_of_divisors++;
        }
        else
        {
            number_of_divisors += 2;
        }
    }
    return number_of_divisors;
	   
   }

    public static void main(String args[])  
    {
        Scanner scan = new Scanner(System.in);
	System.out.print("Enter the number : \n");
	long number = scan.nextLong();
        long number_of_divisors = get_number_of_divisors(number);
        
        System.out.print("Number of Divisors of this Number is : ");
        System.out.print(number_of_divisors);
        scan.close();
    }
}

/*
Standard Input and Output

Enter the number :
1000000000
Number of Divisors of this Number is : 100

Enter the number :
4567323
Number of Divisors of this Number is : 16

Time Complexity : O( sqrt(N) )
Space Complexity : O( 1 )

*/