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Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Meta coding round for a software engineer role, two problems back to back. Nothing too exotic but the second one has more moving parts than it looks like on the surface.

Questions Asked (2)

Q1

Given a word and an abbreviation string, determine whether the abbreviation is valid for that word. Digits represent skipped characters, no leading zeros allowed, and digit sequences can't be empty.

Algorithms & Data Structures
Author's notes

Two pointers, one on each string.

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AI HintsAI Generated

Suggested Approach

Use a two-pointer technique to traverse the word and abbreviation simultaneously, parsing digits as skip counts and comparing letters directly. Validate edge cases such as leading zeros and empty digit sequences during parsing.

Pro tip: Clarify whether the abbreviation can contain digits that are part of the word (e.g., word 'a1' and abbr 'a1' where '1' is a digit) and handle them as skip counts, not literal characters. Also, discuss time and space complexity upfront to show efficiency awareness.

1. Clarify rules and edge cases

Confirm with the interviewer that digits represent skipped characters, no leading zeros, and digit sequences cannot be empty. Discuss cases like empty word, empty abbreviation, and consecutive digits.

2. Initialize two pointers

Set pointers i for word and j for abbreviation, both starting at 0. Iterate while both pointers are within bounds.

3. Parse abbreviation character

If abbr[j] is a digit, parse the full number, checking for leading zeros and ensuring the number is positive. Then advance i by that number and j past the digits.

4. Compare letters

If abbr[j] is a letter, compare it with word[i]. If they match, increment both pointers; otherwise, return false.

5. Validate final positions

After the loop, ensure both pointers have reached the end of their respective strings. If not, the abbreviation is invalid.

Key Points to Mention

  • Two-pointer technique for linear time complexity O(n + m).
  • Handling of digit sequences: parse multi-digit numbers, reject leading zeros, and ensure non-empty sequences.
  • Edge cases: empty word, empty abbreviation, abbreviation longer than word, and digits at the end.
  • Time and space complexity analysis: O(n + m) time and O(1) space.
  • Potential pitfalls: treating digits as characters, mishandling leading zeros, and off-by-one errors in pointer advancement.
  • Testing strategy: walk through examples like 'internationalization' -> 'i12iz4n' and 'apple' -> 'a2e'.

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

Q2

Given a reference to one node in a connected undirected graph, return a deep copy of the entire graph.

Algorithms & Data Structures
Author's notes

The hashmap from original node to its clone is the whole problem.

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AI HintsAI Generated

Suggested Approach

Use a graph traversal (DFS or BFS) to visit each node exactly once, maintaining a hash map from original nodes to their copies. For each visited node, create its copy if not already present, then recursively or iteratively clone all its neighbors and connect them to the copy.

Pro tip: Clarify whether the graph can have cycles or self-loops, and mention that the hash map prevents infinite loops and ensures each node is copied only once. Also, discuss the trade-off between DFS (recursive, may hit stack limits) and BFS (iterative, uses queue) based on graph size.

1. Understand the problem and edge cases

Confirm that the graph is connected, undirected, and may contain cycles. Ask about constraints like node count, whether nodes have unique values, and if the graph can be empty (null input).

2. Choose traversal and data structure

Decide between DFS (recursive or iterative) and BFS. Use a hash map (dictionary) to map original nodes to their cloned counterparts, ensuring each node is copied once.

3. Clone nodes and edges

During traversal, for each original node, create a copy if not already in the map. Then for each neighbor, recursively clone it (if needed) and add the cloned neighbor to the copy's neighbor list.

4. Handle base cases and return

If the input node is null, return null. After traversal, return the clone of the starting node from the hash map.

5. Analyze complexity and test

State that time and space complexity are O(N + E) where N is number of nodes and E is number of edges. Walk through a simple example (e.g., two nodes connected) to verify correctness.

Key Points to Mention

  • Use of a hash map to track visited/cloned nodes to avoid infinite loops in cyclic graphs.
  • Choice of DFS vs BFS and its implications (recursion depth vs queue memory).
  • Time and space complexity analysis: O(N + E) time, O(N) space for the map and recursion/queue.
  • Handling of edge cases: null input, single node, self-loops, and disconnected graphs (though problem states connected).
  • Importance of deep copy: creating new nodes and edges, not just copying references.
  • Potential for iterative DFS to avoid stack overflow in very large graphs.

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