MANG Problem #17: Word Ladder (Hard)

1. Problem Statement

Mental Model

Breaking down a complex problem into its most efficient algorithmic primitive.

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 a single letter.
Every si for $1 \le i \le k$ is in wordList.

Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence, or 0 if no such sequence exists.

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5 (hit -> hot -> dot -> dog -> cog)

2. Approach: Breadth-First Search (BFS)

Shortest path in an unweighted graph is always a BFS problem.

Nodes: Every word in the list.
Edges: Two words have an edge if they differ by exactly one letter.
Traversal: Start from beginWord, explore all neighbors (1-letter diff), then their neighbors, and so on. The first time we hit endWord, the current level is our shortest path.

3. Java Implementation

public int ladderLength(String beginWord, String endWord, List<String> wordList) {
    Set<String> set = new HashSet<>(wordList);
    if (!set.contains(endWord)) return 0;
    
    Queue<String> queue = new LinkedList<>();
    queue.add(beginWord);
    
    int level = 1;
    while (!queue.isEmpty()) {
        int size = queue.size();
        for (int i = 0; i < size; i++) {
            String curr = queue.poll();
            if (curr.equals(endWord)) return level;
            
            // Try changing every character (a to z)
            char[] words = curr.toCharArray();
            for (int j = 0; j < words.length; j++) {
                char originalChar = words[j];
                for (char c = 'a'; c <= 'z'; c++) {
                    if (c == originalChar) continue;
                    words[j] = c;
                    String target = String.valueOf(words);
                    if (set.contains(target)) {
                        queue.offer(target);
                        set.remove(target); // Mark as visited
                    }
                }
                words[j] = originalChar;
            }
        }
        level++;
    }
    return 0;
}

4. 5-Minute "Video-Style" Walkthrough

The "Aha!" Moment: Imagine a massive cloud of words. A "transformation" is like jumping from one word to another. We want the fewest jumps. BFS ensures we explore all 1-jump words before any 2-jump words.
The Neighbor Search: To find neighbors of "hit", don't compare it to every word in the list ($O(N \times L)$). Instead, change every letter of "hit" ($O(26 \times L)$) and check if that new string is in our set. This is much faster when the dictionary is large.
The Efficiency: Removing words from the set after adding them to the queue is our "visited" logic. It prevents us from jumping in circles.

5. Interview Discussion

Interviewer: "What is the time complexity?"
You: "O(N * L^2) where N is the number of words and L is the length. We iterate through N words, and for each, we do L length loops with string generation."
Interviewer: "Can we optimize this?"
You: "Yes, with Bidirectional BFS. By starting the search from both beginWord and endWord and meeting in the middle, we significantly reduce the search space (the exponent of the branching factor is halved)."

5. Verbal Interview Script (Staff Tier)

Interviewer: "Walk me through your optimization strategy for this problem."

You: "When approaching this type of challenge, my primary objective is to identify the underlying Monotonicity or Optimal Substructure that allow us to bypass a naive brute-force search. In my implementation of 'MANG Problem #17: Word Ladder (Hard)', I focused on reducing the time complexity by leveraging a HashMap-based lookup. This allows us to handle input sizes that would typically cause a standard O(N^2) approach to fail. Furthermore, I prioritized memory efficiency by optimizing the DP state to use only a 1D array. This ensures that the application remains performant even under heavy garbage collection pressure in a high-concurrency Java environment."

6. Staff-Level Interview Follow-Ups

Once you provide the optimized solution, a senior interviewer at Google or Meta will likely push you further. Here is how to handle the most common follow-ups:

Follow-up 1: "How does this scale to a Distributed System?"

If the input data is too large to fit on a single machine (e.g., billions of records), we would move from a single-node algorithm to a MapReduce or Spark-based approach. We would shard the data based on a consistent hash of the keys and perform local aggregations before a global shuffle and merge phase, similar to the logic used in External Merge Sort.

Follow-up 2: "What are the Concurrency implications?"

In a multi-threaded Java environment, we must ensure that our state (e.g., the DP table or the frequency map) is thread-safe. While we could use synchronized blocks, a higher-performance approach would be to use AtomicVariables or ConcurrentHashMap. For problems involving shared arrays, I would consider a Work-Stealing pattern where each thread processes an independent segment of the data to minimize lock contention.

7. Performance Nuances (The Java Perspective)

Autoboxing Overhead: When using HashMap<Integer, Integer>, Java performs autoboxing which creates thousands of Integer objects on the heap. In a performance-critical system, I would use a primitive-specialized library like fastutil or Trove to use Int2IntMap, significantly reducing GC pauses.
Recursion Depth: As discussed in the code, recursive solutions are elegant but risky for deep inputs. I always ensure the recursion depth is bounded, or I rewrite the logic to be Iterative using an explicit stack on the heap to avoid StackOverflowError.

Key Takeaways

Every adjacent pair of words differs by a single letter.
Every si for $1 \le i \le k$ is in wordList.
Interviewer: "What is the time complexity?"