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Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Meta coding screen, one algorithmic problem, nothing fancy but it'll trip you up if you don't think carefully about the ordering logic.

Questions Asked (1)

Q1

Given a set of numbers, arrange them to form the largest possible number.

Algorithms & Data Structures
Author's notes

Looks dead simple until you realize sorting by value doesn't work.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints (e.g., non-negative integers, return type) and then propose a greedy sorting approach with a custom comparator. Explain that by sorting the numbers as strings using a comparator that compares concatenations (a+b vs b+a), we can arrange them to form the largest number. Finally, discuss handling edge cases like all zeros and analyze time/space complexity.

Pro tip: Mention that the comparator must be transitive to ensure correct sorting, and note that in languages like Python, you can use functools.cmp_to_key to implement it. Also, be prepared to discuss why a simple descending lexicographic sort fails (e.g., '9' vs '34').

1. Clarify constraints and edge cases

Ask about input size, number ranges, whether numbers can be negative, and expected output format (string vs integer). Identify edge cases like all zeros or single element.

2. Propose greedy sorting with custom comparator

Explain that to maximize the concatenated number, we should sort the numbers such that for any two numbers a and b, if a+b > b+a (as strings), then a should come before b.

3. Implement the comparator and sort

Describe how to implement the comparator, e.g., using a custom sort function or converting to strings and using a key. Mention that the comparator is transitive and thus sorting is valid.

4. Handle edge cases and return result

After sorting, concatenate the strings. If the result starts with '0', return '0' (all zeros case). Otherwise, return the concatenated string.

5. Analyze complexity and discuss alternatives

State that time complexity is O(n log n) due to sorting, with O(n) space for the strings. Briefly mention that a naive approach would be to try all permutations (O(n!)) which is infeasible.

Key Points to Mention

  • Custom comparator: compare a+b vs b+a as strings to decide order.
  • Transitivity of the comparator ensures sorting correctness.
  • Edge case: all zeros should return '0' instead of '000...'.
  • Time complexity: O(n log n) for sorting, O(n) space for strings.
  • Naive permutation approach is O(n!) and impractical.
  • Language-specific implementation details (e.g., Python's cmp_to_key, Java's Comparator).

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