Salesforce·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Salesforce SWE interview focused on linked list manipulation, specifically around removing duplicates. The question had multiple layers to it depending on whether the list was sorted or not, and they pushed into a harder variant at the end.

Questions Asked (2)

Given the head of a singly linked list, remove duplicate nodes so each value appears at most once. How would your approach differ if the list is unsorted versus sorted?

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The sorted case is pretty clean, single pass, O(n) time and constant space, just compare each node to the next and skip.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints and edge cases, then present two distinct solutions: one for unsorted lists using a hash set to track seen values, and one for sorted lists using a two-pointer technique to skip duplicates. Compare their time and space complexities, and discuss trade-offs such as memory usage versus simplicity.

Pro tip: Mention that for sorted lists, you can achieve O(1) space by modifying the list in-place, which is often preferred in memory-constrained environments. Also, note that the unsorted approach can be adapted to sorted lists but may be overkill.

1. Clarify requirements and edge cases

Ask if the list is sorted or unsorted, if duplicates should be removed in-place, and if the list can be empty or have one node. Confirm return type (head of modified list).

2. Unsorted list approach

Use a hash set to track seen values. Traverse the list, and for each node, if its value is in the set, remove it by adjusting pointers; otherwise, add it to the set. Time O(n), space O(n).

3. Sorted list approach

Since duplicates are adjacent, use a pointer to compare current node with next. If equal, skip the next node; else, move forward. Time O(n), space O(1).

4. Compare and discuss trade-offs

Highlight that the sorted approach is more space-efficient but requires sorted input. The unsorted approach works for any list but uses extra memory. Mention that sorting first would take O(n log n) time, which may be worse.

5. Handle edge cases and code

Walk through edge cases like empty list, single node, all duplicates. Write clean code with dummy head if needed to simplify removal of head duplicates.

Key Points to Mention

Time and space complexity for both approaches: O(n) time for both, O(n) space for unsorted, O(1) space for sorted.
Use of a hash set for unsorted lists to track seen values.
Two-pointer technique for sorted lists to skip duplicates in-place.
Trade-offs: memory usage vs. simplicity, and the fact that sorting first adds O(n log n) time.
Edge cases: empty list, single node, duplicates at head, all nodes duplicates.
In-place modification vs. creating a new list, and how to handle the head node (e.g., using a dummy node).

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

What if instead of keeping one copy of each duplicate, you had to remove ALL nodes that appear more than once?

Algorithms & Data Structures

Author's notes

This tripped me up more than I expected.

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AI HintsAI Generated

Suggested Approach

First, clarify that this is a variation of the classic 'remove duplicates' problem where we must delete all nodes with duplicate values, not just keep one. Then, propose a two-pass approach using a hash map to count frequencies, followed by a second pass to remove nodes with count > 1, or a one-pass approach with a dummy head and a hash set to track seen values.

Pro tip: Mention edge cases like an empty list, all duplicates, or duplicates at the head, and discuss trade-offs between time and space (e.g., O(n) time with O(n) space vs. O(n^2) time with O(1) space).

1. Clarify the problem

Confirm that we need to remove all nodes that have duplicate values, meaning if a value appears more than once, none of its occurrences should remain.

2. Choose an approach

Decide between a two-pass hash map approach (count frequencies, then remove) or a one-pass approach using a dummy head and a hash set to track seen values.

3. Handle edge cases

Consider cases like an empty list, a list with all duplicates, duplicates at the head, and duplicates at the tail.

4. Implement and test

Write clean code with a dummy head to simplify removal, and walk through examples to verify correctness.

5. Analyze complexity

State the time and space complexity of your solution and discuss potential optimizations or trade-offs.

Key Points to Mention

Use a dummy head node to handle removal of the head node seamlessly.
Two-pass approach: first count frequencies with a hash map, then traverse again to remove nodes with count > 1.
One-pass approach: use a hash set to track seen values and a pointer to the previous node to skip duplicates.
Time complexity: O(n) for both approaches; space complexity: O(n) for hash map/set.
Edge cases: empty list, single node, all nodes duplicates, duplicates at head/tail.
Trade-off: O(n) space vs. O(n^2) time if using nested loops for O(1) space.

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