Explain the algorithm to reverse a linked list in groups of k nodes.

• A. Reverse the entire list in a single pass without considering blocks.
• B. For each block of k nodes, reverse the links within the block; connect blocks with a dummy head; handle leftover nodes.
• C. Move nodes across blocks in a zigzag pattern to achieve grouping.
• D. Only reverse the pointers of the first node in each block.

Answer: (B)

This question tests how to reverse a linked list in fixed-size blocks, so each block of k nodes ends up in reverse order while the blocks themselves stay in their original sequence.

Think of the list as a chain of segments, each containing up to k nodes. For each full segment, you reverse the links inside that segment, but you keep the overall block order the same. A dummy head is usually used to simplify connecting the first block to the rest and to handle edge cases cleanly. The process is to walk through the list, grab the next k nodes as a block, reverse that block in place, then splice it back into the growing list.

The routine looks like this conceptually: start with a pointer to the node before the current block. Identify the next k nodes; if fewer than k remain, stop and leave them as is. Otherwise, reverse the links within that block. Then connect the previously processed part to the new head of the reversed block and connect the tail of the reversed block to the next portion of the list. Move the previous-block pointer to the end of the reversed block and repeat.

Why this is the right approach: it explicitly enforces reversing only complete blocks and preserves the relative order of blocks, while efficiently performing in-place pointer reversals. It also handles leftovers by leaving any fewer-than-k nodes at the end untouched, which matches the common definition of this operation. Using a dummy head makes the initial connections and edge cases seamless.

Why the other ideas don’t fit: reversing the entire list ignores the block boundaries and would mishandle leftovers. trying to zigzag nodes across blocks would break the requirement of keeping blocks in their original sequence. reversing only the first node in each block would not change the block's internal order to achieve the desired grouping.