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The problem

A sensor log keeps its readings in a doubly linked list sorted by value, and repeated readings are noise. The cleanup job must keep one node per distinct value and free the rest, keeping both directions of the list consistent. Because the list is sorted, all copies of a value sit next to each other, so a single forward pass that compares each node with its successor is enough.

Example: 1 <-> 1 <-> 1 <-> 2 <-> 3 <-> 4 → 1 <-> 2 <-> 3 <-> 4.

Function signature

DNode* removeDuplicates(DNode* head);  // keeps the first node of each value; returns the head

Test cases

Input list	Output list	Explanation
1 <-> 1 <-> 1 <-> 2 <-> 3 <-> 4	1 <-> 2 <-> 3 <-> 4	A run at the head
1 <-> 2 <-> 2 <-> 3 <-> 3 <-> 4 <-> 4	1 <-> 2 <-> 3 <-> 4	Runs in the middle and at the tail
5 <-> 5 <-> 5	5	All equal
1 <-> 2 <-> 3	1 <-> 2 <-> 3	Nothing to remove
(empty)	(empty)

(Confirm: the list is sorted in non-decreasing order; removed nodes are freed; prev pointers must stay correct.)

How interviewers phrase it

• "Remove duplicates from a sorted doubly linked list" (GeeksforGeeks).
• The singly-linked version is LeetCode 83; the variant that removes every value that repeats is LeetCode 82.

Clarifying questions to ask

1. Is the list sorted? (Yes, so duplicates are adjacent.)
2. Keep one copy, or remove all copies of repeated values?
3. Should removed nodes be freed?

Follow-ups to expect

1. "Why does sorting matter?"
2. "Why not advance after deleting?"
3. "What extra work does a doubly linked list need?"
4. "How would you remove all copies of repeated values?"
5. "Complexity?"

Related questions

• Delete all occurrences of a key in DLL Hard
• Flattening of LL Hard
• Introduction to Doubly LL Easy
• Insert node before head in Doubly Linked List Easy
• Delete head of Doubly Linked List Easy
• Reverse a Doubly Linked List Medium

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