CareerCup

i think multimap should be used for mapping student to courses and courses to student.a)O(no. of students) b)O(no. of courses) c)O(1) d)O(1)

Yeah right. Do your own homework.

I think we can we have hash within another hash. We can declare something like this

Hash<Integer,Hash<Integer,String>> studentCourse=new Hash<Integer,Hash<Integer,String>>();

Here 1st Integer represents Student ID and second Integer represent Course ID.

how about have 1 list for students another for courses and a map with key as course and value as the list of students enrolled for that course

How about a matrix of courses Vs students. we can have the association marked as X or sth. the access times will be low for both getting the students for a particular course or getting the courses for a particular students. the space is a hurdle if the matrix tends to be sparse.

With tables in a database, this becomes very easy.
Three tables:
Courses(ID, Name, KEY(ID))
Students(ID, Name, KEY(ID))
StudentCourses(StudentID FOREIGN KEY Students(ID), CourseID FOREIGN KEY Courses(ID), KEY (StudentID, CourseID))

We can also represent this as a matrix of bits.
Example:
Students (S1, S2, S3),
Courses (C1, C2, C3, C4)

S1's Courses: C1, C2
S2's Courses: C2, C3
S3's Courses: C4, C1

Matrix can be (Students x Courses):
1 1 0 0
0 1 1 0
1 0 0 1

In this all operations asked in the question are : O(n). Strictly speaking, O(Courses) for course list (full or a student's) and O(Students) for Students list.

We have to have the relationship stored in some place. Since in the question its mentioned as graph, one can visualize the structure as bipartite graph, left nodes are students, right nodes are courses and edges are relations. For implementation, Student object will have an array of pointers to course and Course object will have an array of pointers to Students.

1) Any number divisible by 6 or 9 -> should be divisible by 3
i.e. if a number is divisible by 3 and is greater then 6 -> then blindly it's possible.
For example -> 6,9,12(6+6),15(9+6),18(9+9),21(6+6+9) etc ...

2)17 is a prime number , so we can devide the number by 17 , get the reminder , check if greater then equal to 6 and divisible by 3 .. :)

so code can be like
Check(int number)
{
if(number >= 6 && number mod 3 == 0) return true;
int reminder = number mod 17 ;
if(reminder >=6 && reminder mod 3 ==0) return true;
return false;
}
Moral -> Think logically :)

Tushar Gupta