Monday, 30 July 2018

Prime Number with different approach, 1) Find prime number withing a range 2) Prime number using math.sqrt 3)Prime number using while loop ? Selenium webdriver experienced interview qus/Java /Prime Number qus/Most asked interview qus

 //PRIME Number within a range
public class PrimeNumber {

public static void main(String[] args) {
  // TODO Auto-generated method stub

  PrimeNumber p = new PrimeNumber();
  System.out.println("====method1========");
  p.isPrime_1(10,100);

  System.out.println("===================");

public void isPrime_1(int s1, int s2) {
         int  flag = 0, i, j;
         System.out.println ("The prime numbers in between the entered limits are :");
         for(i = s1; i <= s2; i++)
         {
             for( j = 2; j < i; j++)
             {
                 if(i % j == 0)
                 {
                     flag = 0;
                     break;
                 }
                 else
                 {
                     flag = 1;
                 }
             }
             if(flag == 1)
             {
                 System.out.println(i);
             }
         }
}

//PRIME Number using math.sqrt

public class PrimeNum {

   public static void main(String[] args) 
  {
       Scanner pnc = new Scanner(System.in);
       System.out.print("Enter a number : ");
       int n = pnc.nextInt();
       if (IsPrimeNumber(n)) 
      {
           System.out.println(n + " is a prime number");
       } else 
          {
           System.out.println(n + " is not a prime number");
          }
   }

   public static boolean IsPrimeNumber(int n) {
       if (n <= 1) 
      {
           return false;
       }
       for (int i = 2; i < Math.sqrt(n); i++) 
      {
           if (n % i == 0)
          {
               return false;
           }
       }
       return true;
   }
}

//PRIME number using while loop 

 public class PrimeNumberWhileLoop {

    public static void main(String[] args) {
         Scanner pn = new Scanner(System.in);
                       System.out.print("Enter a number : ");
                       int number = pn.nextInt();
int i = 2; boolean flag = false; while(i <= number/2) { // condition for not a prime number if(number % i == 0) { flag = true; break; } ++i; } if (!flag) System.out.println(number + " :is a prime number."); else System.out.println(number + " :is not a prime number."); } }

6 comments:

  1. In method 1 where is start range and end range initialized?

    ReplyDelete
    Replies
    1. startrange and endrange will be the input as per requirement. If qus is to find prime number between range 10 to 100 then startrange will be 10 and endrange will be 100

      Delete
    2. I dont think it will work.
      We can discuss on other IM

      Delete
    3. I have updated with two input parameters for start range as s1 and end range as s2

      Delete
  2. Not able to undertsand the use of loops in method 2 and 3

    ReplyDelete
    Replies
    1. In 3rd method we are using num/2 as the max limit for while loop because, the smallest multiple that will not make it a prime is 2. If you have checked all the numbers from 0 to num/2, If multiple by 2 is bigger than n, then a multiple of 3 or 4 etc will also be bigger than num.
      So the largest factor for any number num must be <= num/2
      like 10, if you can not divide it from 2-5 it is prime, 6-9 are not going to divide ii.
      num = the input number which we want to check

      In 2nd method we have used math.sqrt, If a number is not a prime, it can be factored into two factors f1 and f2, If f1 and f2 are > the sqrt of the number, f1*f2 would be > the number. So at least one of those factors must be <= to the sqrt of the number. To see if a number is actually prime, we only need to test factors that are <= to the sqrt.

      Delete

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