Word Ladder
Given two words, `beginWord` and `endWord`, and a dictionary `wordList`, find the length of the shortest transformation sequence from `beginWord` to `endWord`.
Why Interviewers Ask This
Salesforce asks this Hard-level graph problem to evaluate a candidate's mastery of Breadth-First Search (BFS) for shortest path discovery. It specifically tests the ability to model implicit graphs where nodes are words and edges represent single-character transformations. Interviewers assess whether you can optimize space complexity using bidirectional search or efficient neighbor generation, reflecting Salesforce's focus on scalable, high-performance data processing solutions.
How to Answer This Question
1. Clarify constraints: Confirm if wordList contains duplicates, case sensitivity rules, and if beginWord is guaranteed to be in the list. This mirrors Salesforce's emphasis on edge-case handling in production code.
2. Model the problem: Explicitly state that this is an unweighted graph traversal where each node is a word and edges connect words differing by one character.
3. Select the algorithm: Propose BFS as the standard approach for finding the shortest path in unweighted graphs. Mention Bidirectional BFS as an optimization strategy to reduce the search space significantly.
4. Discuss implementation details: Explain how to generate neighbors efficiently without checking every dictionary word against every other word, perhaps by using a hash set for O(1) lookups or a Trie structure.
5. Analyze complexity: Clearly articulate time and space complexity, noting that generating all possible variations of a word takes O(L * 26) where L is word length.
6. Walk through an example: Trace the logic with a small input like 'hit' to 'cog' to demonstrate your thought process before writing code.
Key Points to Cover
- Explicitly identifying the problem as an unweighted graph traversal requiring BFS
- Demonstrating knowledge of Bidirectional BFS as a performance optimization
- Explaining efficient neighbor generation to avoid O(N*M) comparisons
- Correctly handling edge cases like unreachable targets or empty dictionaries
- Articulating clear time and space complexity analysis
Sample Answer
To solve the Word Ladder problem at Salesforce, I first recognize this as a classic shortest-path problem in an unweighted graph. The goal is to find the minimum number of steps to transform beginWord into endWord, changing only one letter at a time, with each intermediate step existing in the provided wordList.
My primary approach would be Breadth-First Search (BFS). Since BFS explores nodes level by level, the first time we encounter the endWord, we are guaranteed to have found the shortest transformation sequence. I would start by initializing a queue with the beginWord and a distance counter of 1. I also need a visited set to prevent cycles and redundant processing.
For efficiency, instead of comparing the current word against every word in the dictionary—which would be too slow—I will generate all possible one-letter mutations of the current word. For each mutation, I check if it exists in the wordList and hasn't been visited yet. If it matches the endWord, I return the current distance plus one. Otherwise, I add it to the queue and mark it as visited.
Given the 'Hard' difficulty and Salesforce's scale, I would consider Bidirectional BFS. By running two simultaneous searches from both the beginWord and endWord, we meet in the middle, effectively reducing the branching factor and improving performance significantly. This demonstrates an understanding of optimizing algorithms for large datasets, which is crucial for enterprise applications. Finally, I'd analyze the time complexity as O(M^2 * N), where M is the word length and N is the number of words, ensuring the solution remains performant within typical interview time limits.
Common Mistakes to Avoid
- Using Depth-First Search (DFS) which finds any path but not necessarily the shortest one
- Failing to use a HashSet for the wordList, leading to O(N) lookups inside the loop
- Ignoring the constraint that intermediate words must exist in the dictionary
- Not considering memory usage when storing visited nodes in very large word lists
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