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🪜 Graph: Word Ladder

📝 Problem Description

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 exactly one letter. - Every si for \(1 \le i \le k\) is in wordList. - sk == endWord. Return the number of words in the shortest transformation sequence, or 0 if no such sequence exists.

Real-World Application

This problem models pathfinding in semantic networks or state-space search where each word represents a state and a one-character change is an edge. It is useful in natural language processing (NLP) for measuring word similarity and in recommendation systems for finding related items.

🛠️ Constraints & Edge Cases

  • \(1 \le\) length of beginWord \(\le 10\)
  • \(1 \le\) wordList.length \(\le 5000\)
  • Edge Cases to Watch:
    • endWord not in wordList.
    • No transformation path exists.

🧠 Approach & Intuition

The Aha! Moment

Instead of comparing every word with every other word to check if they differ by one character (\(O(N^2 \cdot L)\)), use generic patterns (e.g., h*t for hot, dot, lot) as intermediary nodes in the graph. This reduces the edge-building process significantly.

🐢 Brute Force (Naive)

Creating an adjacency matrix where an edge exists if words differ by one character is too slow due to the \(O(N^2)\) comparisons.

🐇 Optimal Approach

  1. Create a dictionary (adjacency list) where keys are generic patterns (e.g., h*t) and values are lists of words matching that pattern.
  2. Perform BFS from beginWord to endWord.
  3. At each step, generate all possible patterns for the current word to find all reachable neighbors in one character change.

🧩 Visual Tracing

graph LR
    HOT[hot] --> |h*t| H_T[Generic]
    H_T --> DOT[dot]
    DOT --> |d*t| D_T[Generic]
    D_T --> DOG[dog]

💻 Solution Implementation

(Implementation details need to be added...)

⏱️ Complexity Analysis

  • Time Complexity: \(\mathcal{O}(M \times N^2)\) or \(\mathcal{O}(M^2 \times N)\) depending on implementation (where \(M\) is length of word, \(N\) is number of words).
  • Space Complexity: \(\mathcal{O}(M^2 \times N)\) to store the pattern dictionary.

🎤 Interview Toolkit

  • Harder Variant: Bidirectional BFS: Search from both beginWord and endWord simultaneously to drastically reduce the search space.
  • Alternative Data Structures: The dictionary-based adjacency list is the core optimization here.