0126. Word Ladder II

https://leetcode.com/problems/word-ladder-ii

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 a single letter.

• Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.

• sk == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words [beginWord, s1, s2, ..., sk].

Example 1:

**Input:** beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
**Output:** [["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]]
**Explanation:** There are 2 shortest transformation sequences:
"hit" -> "hot" -> "dot" -> "dog" -> "cog"
"hit" -> "hot" -> "lot" -> "log" -> "cog"

Example 2:

**Input:** beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
**Output:** []
**Explanation:** The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.

Constraints:

• 1 <= beginWord.length <= 5

• endWord.length == beginWord.length

• 1 <= wordList.length <= 1000

• wordList[i].length == beginWord.length

• beginWord, endWord, and wordList[i] consist of lowercase English letters.

• beginWord != endWord

• All the words in wordList are unique.

ac

class Solution {
    // BFS build a directed Graph, better start from endWord, use map to circumvent loop
    // DFS find paths
    public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
        List<List<String>> res = new ArrayList<List<String>>(); // return result
        // build the graph
        Map<String, List<String>> graph = new HashMap<String, List<String>>(); // the graph
        Map<String, Integer> visitedMap = new HashMap<String, Integer>(); // put visited node and it's layer
        Set<String> wordSet = new HashSet<String>(wordList); // check whether dict contains word
        if (!wordSet.contains(endWord)) return res;
        wordSet.add(beginWord);

        //BFS
        Queue<String> q = new LinkedList<String>();
        q.offer(endWord);
        int layer = 0;
        int shortest = Integer.MAX_VALUE; // stop BFS
        while (!q.isEmpty()) {
            layer++;
            if (layer >= shortest) break; // shortest find, stop BFS
            int len = q.size();
            // layer by layer
            for (int i = 0; i < len; i++) {
                String start = q.poll();
                char[] chars = start.toCharArray();
                for (int k = 0; k < start.length(); k++) {
                    // construct new word
                    char old = chars[k];
                    for (char c = 'a'; c <= 'z'; c++) {
                        chars[k] = c;
                        String next = String.valueOf(chars);
                        // check new word in dict
                        if (wordSet.contains(next)) {
                            if (next == beginWord) shortest = layer + 1;
                            if (!visitedMap.containsKey(next)) {
                                // add to graph
                                List<String> tmpList = new ArrayList<String>();
                                tmpList.add(start);
                                graph.put(next, tmpList);
                                visitedMap.put(next, layer+1);
                                // add to queue for next
                                q.offer(next);
                            } else {
                                if (visitedMap.get(next) < layer + 1) {
                                    continue; // loop, give up this word 
                                } else {
                                    graph.get(next).add(start); // add to graph
                                }
                            }
                        }
                    }
                    chars[k] = old;
                }
            }
        }

        List<String> tmpRes = new ArrayList<String>(); // tmp res note for DFS
        if (!graph.containsKey(beginWord)) return res; // beginWord not even in the graph, return
        dfs(graph, beginWord, endWord, tmpRes, res);
        return res;
    }

    private void dfs(Map<String, List<String>> graph, String beginWord, String endWord, List<String> tmpRes, List<List<String>> res) {
        tmpRes.add(beginWord);
        if (beginWord.equals(endWord)) {
            res.add(new ArrayList<String>(tmpRes));
            tmpRes.remove(beginWord);
            return;
        }
        for (String s : graph.get(beginWord)) {
            // tmpRes.add(s);
            dfs(graph, s, endWord, tmpRes, res);
            // tmpRes.remove(s);
        }
        tmpRes.remove(beginWord);
    }
}

class Solution {
    public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
        List<List<String>> res = new ArrayList<List<String>>();
        Set<String> wordSet = new HashSet<String>(wordList);
        // edge case
        if (!wordSet.contains(endWord)) return res;
        wordSet.add(beginWord);
        wordSet.remove(endWord);  // reverse

        Map<String, List<String>> map = new HashMap<>();  // directed graph
        Map<String, Integer> visited = new HashMap<>();
        Queue<String> q = new LinkedList<>();  // BFS
        q.offer(endWord);
        int layer = 0;
        boolean findShortest = false;
        while (!q.isEmpty() && !findShortest) {
            layer++;
            int len = q.size();
            for (int i = 0; i < len; i++) {
                String h = q.poll();
                char[] chars = h.toCharArray();
                for (int j = 0; j < chars.length; j++) {
                    char old = chars[j];
                    for (char newChar = 'a'; newChar <= 'z'; newChar++) {
                        chars[j] = newChar;
                        String newWord = String.valueOf(chars);

                        if (wordSet.contains(newWord)) {
                            if (newWord.equals(beginWord)) findShortest = true;

                            if (!visited.containsKey(newWord)) {
                                map.put(newWord, new ArrayList<String>());
                                map.get(newWord).add(h);
                                q.offer(newWord);
                                visited.put(newWord, layer + 1);
                            } else {
                                if (visited.get(newWord) < layer + 1) continue;  // visited in last layer, don't go back
                                map.get(newWord).add(h);
                            }
                        }

                    }
                    chars[j] = old;
                }
            }
        }

        // build Path
        if (!map.containsKey(beginWord)) return res;
        List<String> note = new ArrayList<String>();
        note.add(beginWord);
        dfs(map, beginWord, note, res);

        return res;
    }
    private void dfs(Map<String, List<String>> map, String beginWord, List<String> note, List<List<String>> res) {
        if (!map.containsKey(beginWord)) {  // meet the end
            res.add(new ArrayList<String>(note));
            return;
        }

        for (String s : map.get(beginWord)) {
            note.add(s);
            dfs(map, s, note, res);
            note.remove(note.size() - 1);
        }
    }
}

/*
build graph backwards
dfs get result

*/