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Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Jun 2026

Summary

Meta SWE interview with a linked list question that looks trivial until you actually sit down and think through all the edge cases.

Questions Asked (1)

Q1

You're given two singly linked lists where each node holds a single character. Write a function that returns true if both lists represent the same string, false otherwise.

Algorithms & Data Structures
Author's notes

Seemed easy at first glance.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem: confirm that the lists represent strings with the same character order, and discuss edge cases like empty lists. Then propose a two-pointer traversal that compares characters node by node, handling unequal lengths, and analyze time and space complexity.

Pro tip: After presenting the straightforward O(n) time, O(1) space solution, mention the trade-off of converting to strings (O(n) space) and why the two-pointer approach is preferred for large lists or memory-constrained environments.

1. Clarify the problem

Ask if the lists are guaranteed to be non-empty, if characters are case-sensitive, and if the strings must be identical in length and order. Confirm that 'same string' means same sequence of characters.

2. Outline the approach

Propose traversing both lists simultaneously with two pointers, comparing characters at each step. If characters differ or one list ends before the other, return false; otherwise, continue until both end.

3. Handle edge cases

Consider empty lists, lists of different lengths, and lists with the same prefix but different suffixes. Explain how the algorithm handles each case.

4. Analyze complexity

State that the time complexity is O(n) where n is the length of the shorter list (or O(min(m,n))), and space complexity is O(1) since only pointers are used.

5. Discuss alternatives and trade-offs

Mention that converting to strings would use O(n) extra space but might be simpler; however, the two-pointer method is more memory-efficient and avoids unnecessary allocations.

Key Points to Mention

  • Two-pointer traversal to compare characters in order
  • Early termination when a mismatch is found or lengths differ
  • Time complexity O(n) and space complexity O(1)
  • Edge cases: empty lists, different lengths, case sensitivity
  • Trade-offs: converting to strings vs. in-place comparison
  • Handling of null/None pointers to avoid errors

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