A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:
si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.sk == endWordGiven two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.
Example 1:
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] Output: 5 Explanation: 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 Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.
Constraints:
1 <= beginWord.length <= 10endWord.length == beginWord.length1 <= wordList.length <= 5000wordList[i].length == beginWord.lengthbeginWord, endWord, and wordList[i] consist of lowercase English letters.beginWord != endWordwordList are unique.The Word Ladder problem can be modeled as a shortest path search in an implicit graph. Each word represents a node, and an edge exists between two words if they differ by exactly one character. Since we want the minimum number of transformations from the beginWord to the endWord, the natural strategy is to use Breadth-First Search (BFS).
Start BFS from the beginWord and explore all possible one-letter transformations that exist in the dictionary. A hash set is typically used for fast lookups and to remove visited words, preventing repeated work. For each step, generate all possible character substitutions and check if they form valid dictionary words.
To further optimize, many implementations use Bidirectional BFS, where the search starts simultaneously from both the start and end words. This dramatically reduces the search space by meeting in the middle.
The standard BFS approach has a time complexity roughly proportional to O(N * L * 26), where N is the number of words and L is the word length. Space complexity depends on the dictionary and BFS queue.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Breadth-First Search (BFS) | O(N * L * 26) | O(N) |
| Bidirectional BFS | O(N * L * 26) but faster in practice | O(N) |
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This approach leverages Breadth First Search (BFS), where each word is a node in the graph, and an edge exists between two nodes if they differ by exactly one letter. The goal is to find the shortest path from beginWord to endWord.
Start BFS from the beginWord, exploring transformations by changing each letter and checking if the new word exists in the set of available words. Polished implementation should consider preprocessing the word list for efficient lookup.
Time Complexity: O(N * m^2), where N is the number of words in the list and m is the length of each word. Each word pair is compared in the worst case.
Space Complexity: O(N), where N is the number of nodes (words) we store in the queue.
1const ladderLength = function(beginWord, endWord, wordList) {
2 const wordSet = new Set(wordList);
3 if (!wordSet.has
This JavaScript solution employs BFS to traverse transformations, exploring words that differ by a single letter. Each step pushes eligible words to the queue and removes them from the valid set for space optimization. The ladder depth increments with each level of breadth explored.
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Yes, Word Ladder is a well-known graph and BFS interview problem frequently discussed in technical interviews. Variations of the problem, including bidirectional BFS optimizations, are common in FAANG-level interview preparation.
BFS is ideal because it explores nodes level by level and guarantees the shortest path in an unweighted graph. Each transformation represents an edge with equal cost, making BFS the most suitable traversal method.
The optimal approach typically uses Breadth-First Search because the problem asks for the shortest transformation sequence. Many optimized solutions use bidirectional BFS, which starts searching from both the start and end words to reduce the search space.
A hash set is commonly used to store the word list for constant-time lookups and to remove visited words. A queue is used to perform BFS level-by-level while exploring valid transformations.