Question 15

A singly linked list with $n$ nodes can be reversed in-place using three pointers (prev, curr, next):

1. Initialize prev = NULL and curr = head.
2. In each iteration, save next = curr->next, reverse the link by setting curr->next = prev, then advance prev = curr and curr = next.
3. The loop visits each node exactly once, performing $O(1)$ pointer assignments per node.

This algorithm requires $O(1)$ auxiliary space and runs in $\Theta(n)$ time in both best and worst cases. Since any algorithm must examine all $n$ nodes to reverse the list, $\Omega(n)$ time is a lower bound, making $\Theta(n)$ the tight worst-case time complexity. This confirms Option (A) is TRUE.