Describe how to rotate a singly linked list to the right by k places.

• A. Detach last k nodes and move to front.
• B. Only adjust head pointer.
• C. Compute length; k mod length; connect tail to head to form a circle; move head to the length - k th node; break the circle by setting new tail.next to NULL.
• D. Reverse entire list k times.

Answer: (C)

Rotating a singly linked list to the right by k places means moving the last k nodes to the front while preserving their order. The reliable way to do this handles several edge cases and updates pointers correctly.

First, walk the list to get its length and the current tail. Then reduce k by taking k mod length, because rotating by a full cycle (length) brings you back to the same list. If the result is zero, you can leave the list as is. Otherwise, connect the tail to the head to form a circular list. The new head will be the node that comes after the (length − k)th node, so the new tail is this (length − k)th node. Break the circle by setting the new tail’s next pointer to NULL. This approach efficiently achieves the rotation in one pass after knowing the length and handles large k correctly, which other options fail to do.