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Word Ladder Problem

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

Write a C++ program to solve the Word Ladder Problem. A word ladder is a sequence of words where each word differs from the previous word by exactly one letter. Given 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"]
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  • Output: 5
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  • Explanation: 
      

    • Step 1: Create a graph where words are connected if they differ by exactly one letter.
      • 

    • Step 2: Perform breadth-first search from beginWord to find the shortest path to endWord.
      • 

    • Step 3: The shortest transformation is "hit" → "hot" → "dot" → "dog" → "cog".
      • 

    • Step 4: Return the length of this sequence, which is 5.
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    •
Example 2
    

  • Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
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  • Output: 0
    • 

  • Explanation: 
      

    • Step 1: Create a graph where words are connected if they differ by exactly one letter.
      • 

    • Step 2: Check if endWord exists in wordList - it doesn't.
      • 

    • Step 3: Since endWord is not in wordList, no valid transformation sequence exists.
      • 

    • Step 4: Return 0 to indicate no valid sequence exists.
      • 

    •
Constraints
    

  • 1 ≤ beginWord.length ≤ 10
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  • endWord.length == beginWord.length
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  • 1 ≤ wordList.length ≤ 5000
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  • All strings consist of lowercase English letters
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  • All words in wordList are unique
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  • Time Complexity: O(M^2 * N) where M is the length of each word and N is the total number of words
    • 

  • Space Complexity: O(M * N)
    •
QueueSetGoldman SachsZomato
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Hint
Solution

Solution Hints

    
  
  • Use a breadth-first search (BFS) approach to find the shortest path
    • 
  
  • Create adjacency rules using character substitutions
    • 
  
  • Keep track of visited words to avoid cycles
    • 
  
  • Use a queue to process words level by level
    • 
  
  • Consider preprocessing the wordList for faster neighbor finding
    •

Steps to solve by this approach:

 Step 1: Convert wordList to a set for O(1) lookups and check if endWord exists in it.
 Step 2: Initialize a BFS queue with the beginWord and set level counter to 1.
 Step 3: For each word in the queue, try changing each character to all 26 lowercase letters.
 Step 4: If a changed word equals endWord, return level+1 (current level plus this transformation).
 Step 5: If a changed word exists in wordSet, add it to the queue and remove it from wordSet to avoid cycles.
 Step 6: Increment level counter after processing all words at the current level.
 Step 7: If queue becomes empty without finding endWord, return 0 (no valid sequence).

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