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Coding questions · Salesforce

Remove duplicates from a singly linked list

This question evaluates understanding of singly linked list data structures and in-place node manipulation to remove repeated values while preserving relative order, testing competencies in pointer management and duplicate-detection.

You are given the head of a singly linked list of integers. Modify the list in place so that it contains only the first occurrence of each value (i.e., remove any node whose value has already appeared earlier in the list).

  • Each node has fields:
    • data (integer)
    • next (reference to next node, or null at the tail)
  • The relative order of the remaining nodes must stay the same.

Return the head of the updated linked list.

Examples

  1. Input: 3 -> 4 -> 3 -> 6
Output: 3 -> 4 -> 6

  2. Input: 3 -> 4 -> 3 -> 2 -> 6 -> 1 -> 2 -> 6
Output: 3 -> 4 -> 2 -> 6 -> 1

Remove duplicates from a singly linked list

You are given the head of a singly linked list of integers. Modify the list in place so that it contains only the first occurrence of each value (i.e., remove any node whose value has already appeared earlier in the list). The relative order of the remaining nodes must stay the same. Return the head of the updated linked list.

Each node has a data (integer) field and a next reference (or null at the tail).

Marshalling for this console: the linked list is represented as an array of the node values in order (the sequence of data fields from head to tail). Your function receives this array and must return the array of values that remain after removing every node whose value already appeared earlier — preserving the original order.

Examples

  1. Input: [3, 4, 3, 6] (i.e. 3 -> 4 -> 3 -> 6)  
Output: [3, 4, 6]

  2. Input: [3, 4, 3, 2, 6, 1, 2, 6]  
Output: [3, 4, 2, 6, 1]

Constraints

  • 0 <= number of nodes <= 10^5
  • -10^9 <= node value (data) <= 10^9
  • The relative order of the remaining nodes must be preserved
  • Keep only the first occurrence of each value

Hints

  1. Walk the list once from head to tail, tracking which values you have already seen in a hash set.
  2. Keep a node only the first time you encounter its value; skip (unlink) any node whose value is already in the set.
  3. A single pass with a set gives O(n) time and O(n) extra space — no need to compare every pair of nodes.