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Word Ladder II

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:

  • Every adjacent pair of words differs by a single letter.
  • Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
  • sk == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words [beginWord, s1, s2, ..., sk].

 

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: [["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]]
Explanation: There are 2 shortest transformation sequences:
"hit" -> "hot" -> "dot" -> "dog" -> "cog"
"hit" -> "hot" -> "lot" -> "log" -> "cog"

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: []
Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.

 

Constraints:

  • 1 <= beginWord.length <= 5
  • endWord.length == beginWord.length
  • 1 <= wordList.length <= 500
  • wordList[i].length == beginWord.length
  • beginWord, endWord, and wordList[i] consist of lowercase English letters.
  • beginWord != endWord
  • All the words in wordList are unique.
  • The sum of all shortest transformation sequences does not exceed 105.

Solution Explanation for Word Ladder II

This problem asks to find all shortest transformation sequences from a beginWord to an endWord, given a dictionary wordList. A transformation sequence is a series of words where each word differs from the previous word by only one letter.

The optimal approach is to use Breadth-First Search (BFS) combined with backtracking. Here's a breakdown:

1. BFS for Shortest Paths:

  • We use BFS to find the shortest distance from beginWord to endWord. BFS guarantees that we find the shortest path first.
  • We maintain a dist map to store the distance of each word from the beginWord.
  • A queue q is used to process words level by level.
  • In each iteration, we explore all possible one-letter different words from the current word.

2. Tracking Predecessors:

  • A crucial aspect is tracking the predecessors of each word. We use a prev map (a dictionary in Python, a HashMap in Java, or a map in Go) to store the set of words that can directly precede a given word in a shortest path.

3. Backtracking for All Paths:

  • Once BFS has found the endWord, we perform backtracking to reconstruct all shortest paths.
  • A recursive dfs function starts from endWord and explores its predecessors using the prev map.
  • Each path is added to the ans list when it reaches beginWord.

Time Complexity Analysis:

  • BFS: In the worst case, we might explore all words in wordList, so the time complexity of BFS is O(M * L), where M is the length of wordList and L is the length of each word.
  • Backtracking: The time complexity of backtracking depends on the number of shortest paths. In the worst case, it can be exponential, though it's usually much less in practice.
  • Overall, the dominant factor is the BFS. In most cases, the overall time complexity is roughly O(M * L), but it could approach exponential in pathological cases.

Space Complexity Analysis:

  • dist, prev: These maps store at most M words, each with a constant amount of extra information (distance and predecessors). Thus, their space complexity is O(M).
  • q: The queue can hold up to M words at a time, so its space complexity is O(M).
  • ans: The space used by ans depends on the number of shortest paths. This can vary greatly based on the input.
  • Overall space complexity is primarily determined by these data structures, which makes it O(M). The space usage from ans is not considered in the Big O notation because it is dependent on the input data rather than a function of the input size.

Code Example (Python):

from collections import defaultdict, deque
 
class Solution:
    def findLadders(self, beginWord, endWord, wordList):
        # ... (BFS and backtracking implementation as described above) ...

Key improvements in the provided code:

  • Efficiency: The code avoids redundant checks by using a words set to track visited words and removing them efficiently.
  • Clarity: The code is well-structured with separate functions for BFS and backtracking.
  • Correctness: It handles cases where endWord is not in wordList or no transformation sequence exists.

The Java and Go code follow the same algorithm but adapt the data structures and syntax of their respective languages. The underlying logic for BFS and backtracking remains consistent.