HARD NC#92 Graphs - BFS
127. Word Ladder
📖 Problem
A transformation sequence from beginWord to endWord using a wordList. Each step changes exactly one letter. Return the number of words in the shortest transformation sequence (including beginWord). Return 0 if impossible.
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
- •Level-order traversal
- •Queue discipline
- •Shortest-step interpretation
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply BFS reasoning pattern
💡 Approach
- → Model as graph where edges connect words differing by one letter
- → Use BFS to find shortest path from beginWord to endWord
- → For each word, try changing each character to a-z
- → Use Set for O(1) lookup and to avoid revisiting
- → Time: O(N * M * 26) where N = wordList.length, M = word length
- → Space: O(N * M)
🛠️ Hints & Pitfalls
Hints
- •Model as graph where edges connect words differing by one letter
- •Use BFS to find shortest path from beginWord to endWord
- •For each word, try changing each character to a-z
Common Pitfalls
- •Use Set for O(1) lookup and to avoid revisiting
- •Time: O(N * M * 26) where N = wordList.length, M = word length
- •Space: O(N * M)
🧪 Test Cases
Test Case 1
Not run Input:
ladderLength('hit', 'cog', ['hot','dot','dog','lot','log','cog']); Expected:
5 Test Case 2
Not run Input:
ladderLength('hit', 'cog', ['hot','dot','dog','lot','log']); Expected:
0 Test Case 3
Not run Input:
ladderLength(beginWord: string, endWord: string, wordList: string[]); Expected:
Computed from hidden reference 📝 Code Editor
📤 Output