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Reddit·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jul 2026

Summary

Reddit coding interview that was basically a word ladder variant with a twist. The problem itself wasn't too bad once I recognized the BFS angle, but the follow-up questions on complexity and large-dictionary optimizations pushed me more than I expected.

Questions Asked (5)

Q1

Given a start word, an end word, and a dictionary of same-length words, find the shortest transformation sequence where each step changes exactly 1 or 2 characters (substitutions only) and every intermediate word must be in the dictionary. Return the path or an empty list if none exists.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

I recognized the word ladder pattern pretty fast but the '1 or 2 characters' twist tripped me up for a minute.

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AI HintsAI Generated

Suggested Approach

Model the problem as a graph where words are nodes and edges connect words differing by 1 or 2 characters. Use BFS from the start word to find the shortest path to the end word, ensuring all intermediate words are in the dictionary. Reconstruct and return the path, or an empty list if no path exists.

Pro tip: Precompute character patterns (e.g., wildcard masks) to efficiently find neighbors, and discuss the trade-off between precomputation time and query speed. Also, clarify edge cases like start == end or missing words in the dictionary.

1. Clarify and Validate Inputs

Confirm that all words are the same length, the dictionary contains only valid words, and handle edge cases such as start or end not in dictionary or start equals end.

2. Model as a Graph

Treat each word as a node and connect words that differ by exactly 1 or 2 characters. Explain that this forms an unweighted graph where BFS finds the shortest path.

3. Optimize Neighbor Generation

Use pattern matching (e.g., replacing each character with a wildcard) to group words and quickly find neighbors, reducing time from O(N^2 * L) to O(N * L^2).

4. Run BFS and Reconstruct Path

Perform BFS from the start word, tracking parent pointers. Once the end word is reached, backtrack to build the transformation sequence.

5. Analyze Complexity and Trade-offs

Discuss time and space complexity, and compare BFS with bidirectional BFS or A* for potential optimizations, noting trade-offs in implementation complexity.

Key Points to Mention

  • BFS guarantees the shortest path in an unweighted graph.
  • Efficient neighbor generation using wildcard patterns (e.g., for 1-character difference, replace each char with '*'; for 2-character difference, use two wildcards).
  • Handling edge cases: start == end, start or end not in dictionary, no path exists.
  • Time and space complexity: O(N * L^2) with pattern precomputation, where N is dictionary size and L is word length.
  • Trade-offs between precomputation and on-the-fly neighbor checking.
  • Potential optimizations: bidirectional BFS, A* with heuristic, or limiting to 1-character difference if 2-character is too broad.

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

Q2

Why would you choose BFS over DFS for finding the shortest path in this word transformation problem?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Answered this fine.

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AI HintsAI Generated

Suggested Approach

Start by clarifying that BFS explores nodes level by level, guaranteeing the shortest path in unweighted graphs like word transformations. Then contrast with DFS, which may find a path but not necessarily the shortest, and explain why BFS's queue-based approach ensures minimal transformations. Conclude by mentioning that BFS is optimal here because each edge represents one transformation, so the first time you reach the target, it's via the fewest steps.

Pro tip: Mention that while BFS is optimal for shortest path in unweighted graphs, it can be memory-intensive; a bidirectional BFS can significantly reduce search space and is often expected in interviews at top companies like Reddit.

1. Define the problem

State that the word transformation problem is an unweighted graph where each word is a node and edges connect words differing by one letter. The goal is to find the shortest transformation sequence.

2. Explain BFS properties

Describe how BFS explores nodes in increasing order of distance from the start, ensuring the first time the target is reached, it's via the shortest path.

3. Contrast with DFS

Explain that DFS goes deep along one path and may find a longer path first, requiring exhaustive search to guarantee shortest, which is inefficient.

4. Conclude with optimality

Summarize that BFS is the correct choice because it guarantees the shortest path in unweighted graphs, while DFS does not.

5. Mention optimizations

Optionally, note that bidirectional BFS can improve performance by reducing time and space complexity, showing deeper understanding.

Key Points to Mention

  • BFS explores level by level, ensuring shortest path in unweighted graphs.
  • DFS may find a path but not necessarily the shortest; it requires exploring all paths to guarantee shortest.
  • The word transformation problem is modeled as an unweighted graph where each edge has unit weight.
  • BFS uses a queue (FIFO) to track nodes, while DFS uses a stack (LIFO) or recursion.
  • Time complexity: BFS is O(V+E), which is optimal for this problem.
  • Bidirectional BFS can reduce search space by meeting in the middle, improving efficiency.

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

Q3

What is the time and space complexity of your solution?

Algorithms & Data Structures
Author's notes

I fumbled this a bit.

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AI HintsAI Generated

Suggested Approach

Walk through your solution step by step, identifying the dominant operations and how they scale with input size. State the time and space complexity clearly, then briefly justify each with reference to your code or algorithm. If applicable, mention trade-offs and optimizations you considered.

Pro tip: Always relate complexity to the actual constraints (e.g., input size limits) and discuss whether your solution meets them; this shows you think about practical performance, not just theoretical Big-O.

1. Identify input size variables

Define what n, m, etc. represent in your problem (e.g., array length, string length, number of nodes). This sets the context for complexity analysis.

2. Analyze time complexity

Break down your algorithm into loops, recursion, or operations. Determine how many times each operation executes relative to input size, and sum them to get the overall time complexity.

3. Analyze space complexity

Consider all extra space used: data structures, recursion stack, temporary variables. Express it in terms of input size, ignoring constant factors.

4. Justify and simplify

Explain why the complexity is what it is, and simplify to Big-O notation by dropping constants and lower-order terms.

5. Discuss trade-offs and optimizations

Mention if you could trade time for space or vice versa, and whether your solution is optimal or if there's room for improvement.

Key Points to Mention

  • Define variables clearly (e.g., n = number of elements, m = number of edges).
  • Differentiate between average, best, and worst-case complexities if relevant.
  • Account for hidden costs like string concatenation, list resizing, or hash collisions.
  • Include space used by recursion call stack in recursive solutions.
  • Relate complexity to problem constraints to show practical awareness.
  • Acknowledge if your solution is not optimal and suggest potential improvements.

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

Q4

How would you optimize this solution for very large dictionaries?

Algorithms & Data StructuresTechnical Trade-offsSystem Design
Author's notes

This is where it got interesting.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the constraints: what 'very large' means (e.g., millions or billions of entries), memory limits, and performance requirements. Then discuss algorithmic and data structure optimizations, such as using tries, hash maps with open addressing, or external sorting, and trade-offs between time and space. Finally, mention system-level techniques like sharding, compression, or using disk-based storage if needed.

Pro tip: Reddit deals with massive scale, so emphasize practical trade-offs and real-world constraints (e.g., memory vs. latency) rather than just theoretical optimizations. Show awareness of distributed systems and how to handle skewed data.

1. Clarify requirements and constraints

Ask about the size of the dictionary, expected operations (lookup, insert, delete), memory limits, and latency requirements. This shows you don't jump to solutions without understanding the problem.

2. Analyze current solution's bottlenecks

Identify where the current solution fails at scale: memory usage, time complexity, or I/O. For example, a hash map may have high memory overhead due to pointers and load factor.

3. Propose data structure optimizations

Suggest alternatives like tries (for prefix searches), open addressing hash tables (to reduce memory), or succinct data structures (e.g., Bloom filters for membership). Discuss trade-offs.

4. Consider system-level and distributed approaches

If the dictionary is too large for one machine, discuss sharding, consistent hashing, or using external storage (e.g., SSTables, LSM trees). Mention caching hot entries.

5. Evaluate trade-offs and choose the best approach

Summarize the trade-offs (time vs. space, complexity vs. maintainability) and recommend a solution based on the constraints. Be ready to justify your choice.

Key Points to Mention

  • Time and space complexity of different data structures (e.g., hash tables, tries, B-trees)
  • Memory optimization techniques: open addressing, compression, succinct data structures
  • Distributed systems concepts: sharding, consistent hashing, replication
  • External memory algorithms: sorting, LSM trees, SSTables
  • Caching strategies for hot data (e.g., LRU cache)
  • Real-world trade-offs: latency vs. throughput, memory vs. disk, complexity vs. performance

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

Q5

How does your solution behave when the start word is not present in the dictionary?

Algorithms & Data Structures
Author's notes

Short edge case discussion.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem context: this is likely about a word ladder or similar graph traversal where the start word must be in the dictionary. Then, explain that if the start word is not in the dictionary, the solution should immediately return an empty result or indicate no path exists, as the start word is invalid. Emphasize that this check should be done upfront to avoid unnecessary computation.

Pro tip: Mention that you would also validate the end word and consider edge cases like empty dictionary or start equals end, showing thoroughness. Additionally, discuss how this check integrates with the overall algorithm's time complexity.

1. Clarify the problem

Confirm that the question refers to a word ladder or similar problem where the start word must be in the dictionary. Ask if the dictionary is a set for O(1) lookups.

2. Identify the check

State that the first step in the algorithm should be to verify if the start word exists in the dictionary. If not, return an appropriate result (e.g., empty list, -1, or false).

3. Explain the behavior

Describe that without the check, the algorithm might incorrectly proceed or fail; with the check, it gracefully handles the invalid input by short-circuiting.

4. Discuss implications

Mention that this check adds O(1) time (if dictionary is a hash set) and prevents unnecessary BFS/DFS traversal, improving efficiency.

5. Consider edge cases

Bring up related edge cases: end word not in dictionary, start equals end, empty dictionary, and how they should be handled similarly.

Key Points to Mention

  • Early validation of start word in dictionary
  • Return value: empty list, -1, or false depending on problem specification
  • Time complexity: O(1) check with hash set
  • Avoiding unnecessary graph traversal
  • Handling other edge cases like end word not in dictionary
  • Clear communication of assumptions and problem constraints

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