What is the time complexity and auxiliary space complexity of reversing a singly linked list iteratively?

• A. Time: O(n); Space: O(1).
• B. Time: O(n^2); Space: O(1).
• C. Time: O(n); Space: O(n).
• D. Time: O(1); Space: O(1).

Answer: (A)

Reversing a singly linked list iteratively is done by walking through the list once and flipping each node’s next pointer to the previous node as you go. Because you visit each of the n nodes exactly one time and do a constant amount of work per node, the total time grows linearly with the number of nodes: O(n).

For space, you only keep a few references (such as previous, current, and a next reference) during the pass. No new nodes are created and no extra data structures that scale with n are required, so the auxiliary space is constant: O(1). The original nodes remain; the operation is in-place.

Edge cases like an empty list or a single-element list still follow the same rule: time scales with n (which is 0 or 1 in those cases) and the extra space stays constant. This combination—linear time and constant auxiliary space—matches the described option.