Answer Posted / 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`.

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