MANG Problem #28: Word Ladder II (Hard)

1. Problem Statement

Given two words, beginWord and endWord, and a dictionary wordList, return all the shortest transformation sequences from beginWord to endWord.

Each sequence should be a list of words [beginWord, s1, s2, ..., sk] where every adjacent pair differs by one letter.

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

2. Approach: BFS (Distances) + DFS (Backtracking)

The core challenge is finding all shortest paths, not just one.

1. BFS for Distance: Perform a BFS starting from beginWord to calculate the shortest distance from the source to every word in the dictionary. Store this in a Map<String, Integer> distanceMap.
2. Backtracking for Paths: Start from endWord (or beginWord) and use the distanceMap to guide your search.
   • If you are at word W with distance D, only move to neighbors whose distance is D - 1.
   • This ensures every path you find is guaranteed to be a shortest path.

3. Java Implementation

public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
    List<List<String>> res = new ArrayList<>();
    Set<String> dict = new HashSet<>(wordList);
    if (!dict.contains(endWord)) return res;
    
    Map<String, Integer> distances = new HashMap<>();
    Map<String, List<String>> adj = new HashMap<>();
    
    // 1. BFS to find minimum distances and build adjacency graph
    bfs(beginWord, endWord, dict, distances, adj);
    
    // 2. DFS to reconstruct paths
    List<String> path = new ArrayList<>();
    path.add(beginWord);
    dfs(beginWord, endWord, distances, adj, path, res);
    
    return res;
}

private void bfs(String start, String end, Set<String> dict, 
                 Map<String, Integer> dist, Map<String, List<String>> adj) {
    Queue<String> q = new LinkedList<>();
    q.offer(start);
    dist.put(start, 0);
    
    for (String s : dict) adj.put(s, new ArrayList<>());
    adj.put(start, new ArrayList<>());

    boolean found = false;
    while (!q.isEmpty()) {
        int size = q.size();
        for (int i = 0; i < size; i++) {
            String curr = q.poll();
            int d = dist.get(curr);
            
            List<String> neighbors = getNeighbors(curr, dict);
            for (String n : neighbors) {
                adj.get(curr).add(n);
                if (!dist.containsKey(n)) {
                    dist.put(n, d + 1);
                    if (n.equals(end)) found = true;
                    else q.offer(n);
                }
            }
        }
        if (found) break;
    }
}

4. 5-Minute "Video-Style" Walkthrough

1. The "Aha!" Moment: Word Ladder I is about finding a number. Word Ladder II is about Tracing the Map. You cannot find the paths in the BFS pass alone without consuming massive memory.
2. The Guide: The distanceMap is your GPS. It tells you exactly which way to go at every crossroads to stay on the shortest path.
3. The Reverse: Why backtrack? Because it prunes all dead ends. Every step you take using the distanceMap is guaranteed to lead closer to the destination.

5. Interview Discussion

• Interviewer: "Why is this harder than Word Ladder I?"
• You: "In I, we just return the level. In II, we must explore all branches of the same level. If 'hot' and 'dot' both lead to 'dog' in the same number of steps, we need to record both."
• Interviewer: "Complexity?"
• You: "The time complexity is dominated by the number of possible shortest paths, which can be exponential in the worst case. The BFS part is $O(N \times L^2)$."

Read Next

• MANG Problem #29: Burst Balloons (Interval DP)
• MANG Problem #30: Smallest Range Covering K Lists 埋