Salesforce·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Salesforce SWE interview, technical phone screen focused on tree/recursion problems. The main question was a JSON flattening exercise that sounds trivial until you actually have to handle all the edge cases on the spot.

Questions Asked (1)

Given a nested dictionary (where values can be primitives or arrays), write a function that flattens it into a single-level dictionary with dot-separated keys representing the path to each leaf. Walk through your DFS approach, then discuss how you'd handle arrays and edge cases like empty objects or key collisions.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I got the basic recursive case pretty fast, joining keys with dots as you go deeper.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and defining the output format, then walk through a recursive DFS that builds dot-separated keys as it traverses nested dictionaries and arrays. Discuss how you handle arrays (index-based keys), empty objects (skip or include as leaf), and key collisions (detect and resolve).

Pro tip: Mention that you'd use an iterative stack-based DFS to avoid recursion depth limits for deeply nested structures, and that you'd validate the input to ensure it's a dictionary.

1. Clarify requirements and edge cases

Ask about the expected output format, how arrays should be represented (e.g., dot-separated indices), and how to handle empty objects, null values, and key collisions.

2. Design the recursive DFS approach

Explain that you'll traverse the dictionary recursively, maintaining a prefix for the current path, and when a leaf is reached, add the key-value pair to the result.

3. Handle arrays and nested structures

For arrays, iterate over elements and treat indices as part of the key (e.g., 'a.0.b'), and for nested dictionaries, recurse with an updated prefix.

4. Address edge cases and collisions

Discuss how to handle empty objects (skip or include as leaf), key collisions (e.g., if a key already exists, decide to overwrite, throw an error, or append a suffix), and non-dictionary inputs.

5. Analyze complexity and trade-offs

State that time complexity is O(N) where N is the total number of nodes, and space complexity is O(D) for recursion depth; mention iterative alternative to avoid stack overflow.

Key Points to Mention

Recursive DFS with prefix accumulation
Array handling: use indices as keys (e.g., 'arr.0.key')
Empty objects: decide whether to include as leaf or skip
Key collisions: detect and resolve (e.g., overwrite, error, or suffix)
Time and space complexity analysis
Iterative stack-based alternative for deep nesting

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