My linked list fundamentals are solid so I got through the main logic fairly fast.
First, detect cycles in both lists using Floyd's cycle-finding algorithm. Then, based on the presence of cycles, find the entry points and use two-pointer techniques to determine if the lists intersect.
Pro tip: Clarify with the interviewer whether the lists are singly linked and if modifying the lists is allowed; this shows attention to constraints and can simplify the solution.
Use Floyd's tortoise and hare algorithm to determine if each list has a cycle. If a cycle is found, also find the entry point of the cycle.
If neither list has a cycle, find the lengths of both lists, align the starting points, and traverse to check for a common node.
If only one list has a cycle, they cannot intersect because a cycle would make the other list also have a cycle if they shared a node.
If both have cycles, check if they share the same cycle entry point. If not, traverse the cycle of one list to see if the other's entry point is encountered.
Return true if a common node is found, otherwise false.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the problem: are we dealing with two singly linked lists? If so, the goal is to find the node where they merge. Use a two-pointer technique: traverse both lists, and when one pointer reaches the end, redirect it to the head of the other list. If they intersect, the pointers will meet at the intersection node after at most two passes; if not, both will become null simultaneously.
Pro tip: Mention that this approach works in O(m+n) time and O(1) space, and that it elegantly handles lists of different lengths without explicitly computing their lengths. Also, note that if the lists might have cycles, you'd need to detect and handle that first.
Confirm that the input consists of two singly linked lists that may intersect, and that we need to return the intersecting node (or null if none). Ask if the lists can have cycles or if they are guaranteed acyclic.
Decide between the two-pointer switching technique (O(1) space) or using a hash set to store visited nodes (O(n) space). For optimal space, prefer the two-pointer method.
Initialize two pointers at the heads of the lists. Traverse both simultaneously, advancing each pointer one step at a time. When a pointer reaches the end, redirect it to the other list's head. If the lists intersect, the pointers will meet at the intersection node; if not, they will both become null after at most m+n steps.
State that time complexity is O(m+n) and space is O(1). Discuss edge cases: one or both lists empty, no intersection, intersection at the head, or lists of vastly different lengths.
Walk through a concrete example, such as list A: 1->2->3->4->5 and list B: 6->7->4->5, where the intersection is at node 4. Show how pointers move and meet.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.