Singly Linked List — Remove Duplicate Values

Online assessment classic asked in Salesforce's HackerRank OA, both SMTS and MTS levels. Given a singly linked list, keep only the first occurrence of each value and remove every later node with a value seen earlier. Return the head of the modified list. Requirements Input: head of a singly linked list of integers. Operate on the list (in-place pointer rewiring is fine; allocating a new list also accepted unless the prompt forbids it). Keep only the first occurrence of each value; remove every subsequent duplicate. Return the head of the modified list. Node definition: LinkedListNode { int data; LinkedListNode next; }. Examples Input: 3 -> 4 -> 3 -> 6 Output: 3 -> 4 -> 6 Input: 3 -> 4 -> 3 -> 2 -> 6 -> 1 -> 2 -> 6 Output: 3 -> 4 -> 2 -> 6 -> 1 Notes O(n) hash-set walk is the expected solution: iterate with a (prev, curr) pointer pair, keep a seen set, and when curr.data ∈ seen, splice out curr by setting prev.next = curr.next; otherwise advance prev = curr. O(n²) two-pointer scan (no extra memory) is acceptable if the interviewer constrains memory — for each retained node, walk forward and skip any later node matching its value. Mention the trade-off explicitly. Edge cases: empty list (head == null), single node, all-distinct list (output identical), all-duplicate list (output is one node), duplicates clustered at the head. Take care to not dereference curr.next after splicing — advance via the stored next pointer, not prev.next. Do not use the adjacent-duplicate shortcut for a sorted list: this input is unsorted, so the implementation needs a global seen set or a full suffix scan. Preparation Implement both the O(n)/O(n) hash-set version and the O(n²)/O(1) two-pointer version. Practise the splice logic by hand on a 4-node list where the duplicate sits at the tail. Be ready for the in-OA follow-up: "remove all occurrences of a duplicated value". The structure changes — you need a two-pass count pre-pass or a recursion that returns whether the current value is duplicated.