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Knew right away it was a hash set problem.
Use a hash set to track seen values while traversing the list with two pointers (current and previous). When a duplicate is found, adjust the previous node's next pointer to skip the duplicate; otherwise, add the value to the set and move forward. Return the head of the modified list.
Pro tip: Clarify whether the list is sorted or not, as it affects the optimal approach. Also, discuss edge cases like empty list, single node, and all duplicates to show thoroughness.
Restate the problem to ensure clarity: remove duplicates keeping first occurrence, return head. Ask clarifying questions about list properties (sorted? memory constraints?).
Decide on using a hash set for O(1) lookups to track seen values. Consider trade-offs: O(n) time and O(n) space vs. O(n^2) time and O(1) space if no extra space allowed.
Outline traversal with two pointers: 'prev' and 'current'. For each node, check if its value is in the set; if yes, skip it by updating prev.next; else, add to set and advance prev.
Consider empty list, single node, duplicates at head, and all nodes duplicates. Ensure code handles these without errors.
State time complexity O(n) and space complexity O(n) due to hash set. Mention alternative if no extra space allowed: O(n^2) time with nested loops.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.