Shortest Word Distance
Given an array of strings `wordsDict` and two words `word1` and `word2`, return the shortest distance between the two words in the list.
Why Interviewers Ask This
Spotify asks this to evaluate your ability to optimize space complexity while maintaining linear time performance. They value engineers who can solve problems efficiently without unnecessary memory overhead, reflecting their need for scalable audio streaming infrastructure where resource management is critical.
How to Answer This Question
1. Clarify constraints: Ask if the list contains duplicates or if word1 and word2 are guaranteed to exist.
2. Propose a single-pass solution: Explain that you will track indices of both words using two variables instead of storing all positions in a map.
3. Initialize pointers: Set both index trackers to -1 to indicate they haven't been found yet.
4. Iterate through the array: For each word, update the corresponding tracker and calculate the distance only if both have been seen.
5. Update minimum: Compare current distance with the running minimum and store the smaller value.
6. Return result: Output the final minimum distance after one complete traversal, emphasizing O(1) space usage.
Key Points to Cover
- Demonstrating O(1) space complexity optimization rather than using a hash map
- Handling the logic of updating the minimum distance only when both words are found
- Explaining the single-pass iteration strategy clearly
- Addressing edge cases like duplicate words or missing inputs
- Connecting the solution to real-world scalability needs
Sample Answer
To solve the Shortest Word Distance problem efficiently, I would start by clarifying edge cases, such as whether the input words are always present. Assuming standard conditions, my approach focuses on minimizing space complexity to O(1), which aligns with Spotify's emphasis on scalable systems. I propose a single-pass algorithm using two integer variables to track the most recent indices of word1 and word2. As I iterate through the wordsDict array, whenever I encounter word1, I update its index variable; similarly for word2. Crucially, before updating either index, I check if the other word has already been encountered. If both indices are valid, I calculate the absolute difference between them and compare it against my current minimum distance. If this new distance is smaller, I update the minimum. This ensures we capture the shortest gap dynamically without needing to store multiple positions for each word. For example, in an array like ['apple', 'banana', 'apple', 'orange'], finding the distance between 'apple' and 'banana' requires checking indices 0 and 1, then later 2 and 1, updating the minimum accordingly. This method runs in O(n) time because we traverse the list once, and uses constant space since we only store two integers regardless of input size. This balance of speed and memory efficiency is exactly what I aim for when designing algorithms for high-throughput environments.
Common Mistakes to Avoid
- Storing all indices for each word in a list, which increases space complexity to O(n)
- Failing to handle the case where word1 or word2 appears multiple times in the array
- Calculating distance before confirming both words have been encountered at least once
- Using nested loops resulting in O(n^2) time complexity unnecessarily
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