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Salesforce·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Apr 2026

Summary

Salesforce coding round, one question on linked lists. Pretty standard stuff but the implementation details can trip you up if you're not careful.

Questions Asked (1)

Q1

Given a singly linked list of integers, remove all duplicate nodes so that only the first occurrence of each value is kept. Return the head of the modified list.

Algorithms & Data Structures
Author's notes

Knew right away it was a hash set problem.

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AI HintsAI Generated

Suggested Approach

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.

1. Understand the problem

Restate the problem to ensure clarity: remove duplicates keeping first occurrence, return head. Ask clarifying questions about list properties (sorted? memory constraints?).

2. Choose data structures

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.

3. Design algorithm

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.

4. Handle edge cases

Consider empty list, single node, duplicates at head, and all nodes duplicates. Ensure code handles these without errors.

5. Analyze complexity

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.

Key Points to Mention

  • Use of hash set for O(1) lookups to track seen values.
  • Two-pointer technique (prev and current) to modify links.
  • Time and space complexity analysis: O(n) time, O(n) space.
  • Edge cases: empty list, single node, duplicates at head, all duplicates.
  • Alternative approach if no extra space: O(n^2) time with nested loops.
  • Clarify if list is sorted; if sorted, can use O(1) space by comparing adjacent nodes.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.