Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Meta SWE coding question, pretty straightforward dictionary manipulation problem. Nothing too wild but the efficiency angle made me think twice about how I was sorting.

Questions Asked (1)

Given a dictionary, write a function that iterates over its keys in ascending order and returns the corresponding values in that order.

Algorithms & Data Structures

Author's notes

My first instinct was to just loop through the dict directly, which obviously doesn't give you sorted order.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints (e.g., key type, mutability, return format) and then propose a solution that sorts the keys and retrieves values in that order. Discuss time and space complexity, and consider edge cases like empty dictionaries or non-comparable keys.

Pro tip: Mention that if the dictionary is large and keys are sortable, sorting keys is O(n log n), but if keys are already sorted or can be bucketed, you might achieve O(n). Also, note that in Python 3.7+ dictionaries preserve insertion order, but that doesn't guarantee sorted order.

1. Clarify requirements

Ask about key types, whether keys are comparable, and if the dictionary can be modified. Confirm the expected return type (e.g., list of values).

2. Outline approach

Explain that you will extract keys, sort them, then iterate to collect values. Mention alternative approaches if keys are not sortable.

3. Analyze complexity

State time complexity O(n log n) due to sorting and space complexity O(n) for the sorted keys and result list.

4. Handle edge cases

Discuss empty dictionary, single key, duplicate keys (not possible in dict), and non-comparable keys (e.g., mixed types).

5. Write code

Implement the function in a clean, readable manner, using built-in sorting and list comprehension if appropriate.

Key Points to Mention

Sorting keys: sorted(d.keys()) or sorted(d) in Python.
Time complexity: O(n log n) for sorting, O(n) for iteration.
Space complexity: O(n) for sorted keys and result list.
Edge cases: empty dict, non-comparable keys (e.g., int and str).
Alternative: if keys are integers in a known range, use counting sort for O(n).
Language-specific: In Python, use sorted(d.items()) to get key-value pairs sorted by key.

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