Why do you align two lists by length before checking for intersection by reference?

• A. To ensure you compare nodes at the same distance from the tail; O(n)
• B. To sort the lists first
• C. To minimize memory usage
• D. To maximize the number of traversals

Answer: (A)

When checking two singly linked lists for an intersection by reference, aligning by length ensures you compare nodes that are the same distance from the end of each list. If one list is longer, its extra nodes would offset a parallel walk, so you advance the pointer on the longer list by the difference in lengths. After that, moving both pointers forward in tandem lets you detect the intersection exactly when they point to the same node by reference. If you reach the end without a shared node, there is no intersection. This approach is efficient: linear time with constant extra space. Sorting first isn’t relevant to finding a shared node by reference, and simply increasing traversals or trying to minimize memory in other ways doesn’t address the alignment issue.