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My first instinct was a set to track seen values and just relink nodes as I walked the list.
Use a hash set to track seen values while traversing the list with a pointer, removing any node whose value is already in the set. Alternatively, if the list is sorted, use a two-pointer technique to skip duplicates in O(1) space. Clarify with the interviewer whether the list is sorted and discuss trade-offs between time and space.
Pro tip: Always clarify if the list is sorted; if not, a hash set is optimal, but mention that sorting first would change the order and is not allowed. Also, handle edge cases like empty list or single node upfront.
Ask if the list is sorted, if duplicates are based on value only, and if we can use extra space. Confirm that order must be preserved and removal is in-place.
If sorted, use two pointers to skip duplicates in O(n) time and O(1) space. If unsorted, use a hash set to track seen values, requiring O(n) space.
For hash set: initialize a dummy node pointing to head, traverse with prev and curr. If curr.val in set, skip curr by prev.next = curr.next; else add to set and move prev. For two pointers: use current and runner to skip duplicates.
Consider empty list, single node, all duplicates, and duplicates at head or tail. Ensure the head is updated correctly if the first node is removed.
State time and space complexity. Walk through a small example to verify correctness, and mention potential follow-ups like removing duplicates from an unsorted list without extra space (not possible in O(n) time).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.