Answer / Ashutosh Upadhyay

Here's an iterative solution for counting nodes in a circular linked list. Let `slow` and `fast` be pointers.nn1. Initialize `slow` to the head of the list, and `fast` to `NULL`.n2. Move `fast` two steps ahead of `slow` (i.e., `fast = slow->next->next`).n3. Continue moving `slow` one step ahead (i.e., `slow = slow->next`).n4. If `fast` and `slow` meet, the number of nodes in the circular linked list is one less than the number of steps taken by `slow` since they would have met at the starting node.n5. After step 4, move `slow` to the beginning of the list (i.e., `slow = head`) and increment a counter for each step taken until meeting the original position of `fast`.

How do you sort a map by key?

1 Answers  

---

How can one find a cycle in the linked list? IF found how to recognize the cycle and delete that cycle?

12 Answers   Citrix,

---

State the difference between primitive and non-primitive data types?

1 Answers  

---

What method removes the value from the top of a stack?

1 Answers  

---

Can you override methods of arraylist?

1 Answers  

---

Define red-black trees.

1 Answers   Wipro,

---

what is hashing

7 Answers   Infosys, Nirma,

---

What is a Stack? Explain with example?

1 Answers  

---

What is fibonacci search?

1 Answers  

---

How do you reference all the elements in a one-dimension array?

1 Answers  

---

What happens when arraylist is full?

1 Answers  

---

What is significance of ” * ” ?

1 Answers  

---