To solve this problem, use a breadth-first search (BFS) with a hash table to track transformations from the start word to the end word. The BFS ensures the shortest path is found, and the hash table handles word lookups efficiently. If no path exists, return 0.
Problem Statement
Given a start word beginWord, an end word endWord, and a list of words wordList, the task is to find the number of words in the shortest transformation sequence from beginWord to endWord. Each transformation involves changing exactly one letter, and the new word must exist in wordList. Return the length of the sequence or 0 if no such transformation exists.
For example, given beginWord = "hit", endWord = "cog", and wordList = ["hot","dot","dog","lot","log","cog"], the sequence is "hit" -> "hot" -> "dot" -> "dog" -> "cog", and the result is 5. If the end word is not present in the list, or if no valid sequence exists, return 0.
Examples
Example 1
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.
Example 2
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: 0
The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.
Constraints
- 1 <= beginWord.length <= 10
- endWord.length == beginWord.length
- 1 <= wordList.length <= 5000
- wordList[i].length == beginWord.length
- beginWord, endWord, and wordList[i] consist of lowercase English letters.
- beginWord != endWord
- All the words in wordList are unique.
Solution Approach
Breadth-First Search (BFS)
BFS is used to explore all possible transformations from the beginWord. Starting with beginWord, transform one letter at a time and check if the resulting word is in wordList. Each valid transformation is enqueued for further exploration, and we track the number of transformations. The BFS ensures that the first time we reach the endWord, it is via the shortest transformation sequence.
Hash Table for Word Lookup
A hash table (or set) is used to store the wordList, enabling quick lookups for valid transformations. This speeds up the process of checking if a word exists in the list, which is crucial in BFS. By using the hash table, we avoid repeatedly scanning the entire list, ensuring an efficient solution.
Word Transformation Logic
For each word in the BFS queue, generate all possible valid transformations by changing one letter at a time. For example, starting with 'hit', check words like 'hot', 'dot', etc. Each valid word is added to the queue and removed from the hash table to prevent revisiting. This continues until the endWord is found or all possibilities are exhausted.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity depends on the number of words and the length of each word, with BFS exploring all potential transformations. If there are n words, each of length m, the time complexity is O(n * m), where n is the number of words in wordList. The space complexity is O(n) for storing the hash table and BFS queue.
What Interviewers Usually Probe
- Do you understand how BFS guarantees the shortest path in a word transformation sequence?
- Can you explain why a hash table improves the performance of this problem?
- Will you describe how to generate valid word transformations efficiently?
Common Pitfalls or Variants
Common pitfalls
- Not checking if
endWordis in thewordListbefore starting the BFS, leading to unnecessary computation. - Failing to track visited words properly, which can cause infinite loops or revisiting words.
- Misunderstanding the need to change exactly one letter at a time, which can lead to invalid transformations.
Follow-up variants
- Find the shortest transformation sequence from
beginWordtoendWord, but with an added constraint of a maximum transformation length. - Extend the problem to find all shortest transformation sequences from
beginWordtoendWord. - Modify the problem to allow transformations from
beginWordtoendWordwithout the restriction of using the given word list.
How GhostInterview Helps
- Screenshot of problem setup, including input words and the word list.
- Pathfinding through BFS, with step-by-step complexity breakdowns for both time and space.
- Support for real-time screen-sharing of BFS exploration and word transformation flow.
Topic Pages
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- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the core idea behind the BFS approach for solving Word Ladder?
The BFS approach explores all possible word transformations level by level, ensuring the shortest transformation sequence is found as soon as the end word is reached.
Why is a hash table important in the Word Ladder problem?
A hash table allows constant time complexity lookups to check if a transformed word is part of the word list, significantly improving performance during BFS.
How do you handle already visited words in the Word Ladder problem?
Visited words are removed from the hash table as soon as they are processed in BFS, ensuring they are not revisited and preventing infinite loops.
What happens if the end word is not in the word list?
If the end word is not in the word list, the transformation sequence is impossible, and the function should return 0.
How do you optimize the word transformations in Word Ladder?
Word transformations are optimized by generating valid transformations (one letter at a time) and using a hash table to check if each transformation exists in the word list.
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Find all shortest transformation sequences from beginWord to endWord using a dictionary, leveraging backtracking search with pruning for efficiency.
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