The O(n) Club: Word Ladder (LeetCode 127): BFS Wizardry Meets Word Jenga

The O(n) Club: Word Ladder (LeetCode 127): BFS Wizardry Meets Word Jenga

⚡ TL;DR

Turn beginWord into endWord, one letter per move, using only words in a given list—and the shortest path wins. Don’t care about the path, just the total steps. Brute force? Don’t even bother unless you want your laptop to combust. BFS is your lifeline.

Code not to write in production:

// Attempting every sequence: welcome to O(2^n) doom.
public int ladderLength(String beginWord, String endWord, List
<string> wordList) {
    // Please don’t.
    return 0;
}</string>

🧠 How Devs Usually Mess This Up

Great ways to fail this problem:

🔍 What’s Actually Going On

Picture every word as a socially awkward bot at a tech meetup—they’ll only talk to bots that differ by exactly one character. Building this chain of strained small talk? That’s a graph where words are nodes, one-letter swaps are edges, and your goal is to get from beginWord to endWord in as few introductions as humanly (or bot-ly) possible.

BFS (Breadth-First Search) is the efficient way—the same logic behind ordering coffee in a conference line instead of teleporting across the floor plan. Queue, mark visits, repeat until you find your exit.

🛠️ PseudoCode

• If endWord isn’t in the word list, just return 0 and declare defeat early.
  // No destination? Why even start?
• Precompute neighbor patterns: for each word, replace each letter with ‘*’, building a map like “h*t” → [“hot”, …].
  // Pattern: "d*g" → ["dog", "dig"]
• Use BFS: initialize a queue with beginWord; track visited words in a Set so you don’t endlessly ping-pong.
• BFS each layer:
  • For every position in the word, make the *-pattern and fetch neighbors.
  • Add unvisited neighbors to queue, marking them visited so you never revisit your ex-words.
  • If you hit endWord, return the ladder length. Treat yourself to pizza.
• No ladder? Return 0. Maybe Word Sudoku is more your thing.

💻 The Code

public int ladderLength(String beginWord, String endWord, List<string> wordList) {
    Set
<string> dict = new HashSet<>(wordList);
    if (!dict.contains(endWord)) return 0;
     Queue
<string> queue = new LinkedList<>();
    queue.add(beginWord);
    Set
<string> visited = new HashSet<>();
    visited.add(beginWord);
    int wordLen = beginWord.length();
    int steps = 1;
     Map<string list>> patterns = new HashMap<>();
    for (String word : dict) {
        for (int i = 0; i < wordLen; i++) {
            String pattern = word.substring(0, i) + "*" + word.substring(i + 1);
            patterns.computeIfAbsent(pattern, x -> new ArrayList<>()).add(word);
        }
    }
     while (!queue.isEmpty()) {
        int size = queue.size();
        for (int q = 0; q < size; q++) {
            String curr = queue.poll();
            for (int i = 0; i < wordLen; i++) {
                String pattern = curr.substring(0, i) + "*" + curr.substring(i + 1);
                for (String next : patterns.getOrDefault(pattern, new ArrayList<>())) {
                    if (next.equals(endWord)) return steps + 1;
                    if (!visited.contains(next)) {
                        visited.add(next);
                        queue.add(next);
                    }
                }
            }
        }
        steps++;
    }
    return 0;
}</string></string></string></string></string>

⚠️ Pitfalls, Traps & Runtime Smacks

• Bail if endWord isn’t present—you cannot transform into thin air.
• Don’t revisit words: mark as visited when queued (not after dequeue, or hello, loops).
• Gargantuan word lists? Pattern maps turn O(N × L) misery into O(1) lookups.
• Complexity check: time is O(N × L²), space is O(N × L). This isn’t LeetCode Golf, but don’t run it on War and Peace.

🧠 Interviewer Brain-Tattoo

“Word Ladder is like networking for introvert strings—change one letter, never double-dip, and don’t follow the wrong crowd.”