Sunday, 10 June 2012

One to many relationship in hibernate using annotation -2



Today i am going to tell you how to do one to many relationship in hiberantae by creating a intermediate table. Before going through this post first go through the First Post then only you can understand this post. So in the earlier post i expained you how to do one to many relationship without using the intermediate table.
To develop the one to many mapping using intermediate table we have to make few changes in the Student class and course class.






Please make below changes in the student Class-

Here we add one more annotation @JoinTable to make a intermediate table.


For the course class remove the studentId field.





That changes only we have to make.

Wednesday, 6 June 2012

How to find a String is a circular representation of other String

Many Interviewer may ask this question, "How will you find a string is circular of other String." Suppose we have two strings A and B. we have to check or tell that the B is circular of String A or not. A and B can be any string. If we think for a while , we will get the different solutions like using nested loop we can compare each alphabet of B and A. Other way is we can use the linked list. For both approaches time and space complexity would be high. If you told above solution your answer would be right but it is not optimized . Interviewer might be expecting optimize answer, our answer won't be impress to interviewer.






Now we have to think how we can optimize the time and space complexity for a given problem, so our answer can imprint the image of us on mind of interviewer. To do the needful in the optimize way, we have a solution. Solution is


public class CircularString{
public static void main(String[] args) {
String A="ABCDEFG";
String B="FGABCDE";
String C=A+A;//ABCDEFGABCDEFG
boolean status= C.contains(B);
System.out.println(status);
}
}






In the above example complexity would always be 1. Here C represent ABCDEFGABCDEFG. B conatains "FGABCDE". highlight portion of C equals to the String B. If you tell this solution instead of other solution, it will help you to crack the interview.