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Word Ladder II

Certification: Advanced Level Accuracy: 0% Submissions: 0 Points: 15

Write a C program to find all shortest transformation sequences from beginWord to endWord. A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk where: Every adjacent pair of words differs by a single letter, every si for 1 <= i <= k is in wordList, and 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.

Example 1
    
      
  • Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
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  • Output: [["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]]
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  • Explanation: 
    • Start with "hit", can transform to "hot" (change 'i' to 'o'). 
    • From "hot", can go to "dot" or "lot" (change 'h' to 'd' or 'l'). 
    • From "dot" -> "dog" -> "cog" or from "lot" -> "log" -> "cog". 
    • Both paths have length 5, so both are shortest transformation sequences.
    •
Example 2
    
      
  • Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
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  • Output: []
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  • Explanation: 
    • Start with "hit", can transform to "hot". 
    • From "hot", can reach "dot", "lot". 
    • From "dot" can reach "dog", from "lot" can reach "log". 
    • Neither "dog" nor "log" can reach "cog" since "cog" is not in wordList. 
    • Therefore, no transformation sequence exists.
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Constraints
    
      
  • 1 ≤ beginWord.length ≤ 5
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  • endWord.length == beginWord.length
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  • 1 ≤ wordList.length ≤ 500
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  • wordList[i].length == beginWord.length
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  • All words contain only lowercase English letters
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  • Time Complexity: O(N * M * 26) where N is number of words, M is word length
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  • Space Complexity: O(N * M)
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Hint
Solution

Solution Hints

    
      
  • Use BFS to find shortest path length first
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  • Build adjacency graph of words that differ by one character
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  • Use backtracking to find all paths of shortest length
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  • Keep track of visited words to avoid cycles
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  • Store parent relationships during BFS for path reconstruction
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  • Use level-order traversal to ensure shortest paths
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Steps to solve by this approach:

 Step 1: Check if endWord exists in wordList, return empty result if not found
 Step 2: Use BFS to find the shortest transformation length from beginWord to endWord
 Step 3: Create a queue and add beginWord as starting point with level 1
 Step 4: For each word in queue, find all adjacent words (differ by one character) in wordList
 Step 5: Continue BFS until endWord is reached to get shortest path length
 Step 6: Use backtracking to find all possible paths of the shortest length
 Step 7: Return all valid transformation sequences that achieve the shortest length

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