0127. Word Ladder

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

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 the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Example 1:

**Input:** beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
**Output:** 5
**Explanation:** One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.

Example 2:

**Input:** beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
**Output:** 0
**Explanation:** The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.

Constraints:

• 1 <= beginWord.length <= 10

• endWord.length == beginWord.length

• 1 <= wordList.length <= 5000

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

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

• beginWord != endWord

• All the words in wordList are unique.

ac1: Graph + BFS

12/22/2017

very time consuming in building the graph Buiding the graph is awaste of time here.

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        wordList.add(beginWord);
        int begin = wordList.size() - 1;
        int end = -1;

        // Build graph, iterate wordlist get edges.
        List<List<Integer>> adjList = new ArrayList<List<Integer>>();
        for (int i = 0; i < wordList.size(); i++) {
            adjList.add(new ArrayList<Integer>());
        }

        for (int i = 0; i < wordList.size(); i++) {
            if (wordList.get(i).equals(endWord)) end = i;
            for (int j = i + 1; j < wordList.size(); j++) {
                if (isConnected(wordList.get(i), wordList.get(j))) {
                    adjList.get(i).add(j);
                    adjList.get(j).add(i);
                }
            }
        }

        // BFS
        int layer = 0;
        Queue<Integer> q = new LinkedList<Integer>();
        boolean[] visited = new boolean[wordList.size()];
        q.offer(begin);
        while (!q.isEmpty()) {
            int len = q.size();
            layer++;
            for (int i = 0; i < len; i++) {
                int head = q.poll();
                visited[head] = true;
                for (int next : adjList.get(head)) {
                    if (visited[next]) continue;
                    if (next == end) return layer+1;
                    q.offer(next);
                }
            }

        }

        // can't find
        return 0;
    }
    private boolean isConnected(String s1, String s2) {
        int diff = 0;
        for (int i = 0; i < s1.length(); i++) {
            if (s1.charAt(i) != s2.charAt(i)) diff++;
        }
        return diff == 1;
    }
}

ac2: BFS

Simpler and much faster. iterate through word is much faster iterate wordList.

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> wordSet = new HashSet<String>(wordList);

        int layer = 0;

        Queue<String> q = new LinkedList<String>();
        q.offer(beginWord);
        while (!q.isEmpty()) {
            int len = q.size();
            layer++;
            for (int i = 0; i < len; i++) {
                String head = q.poll();
                // get head's adjcent words in dict
                for (int l = 0; l < head.length(); l++) {
                    for (char c = 'a'; c <= 'z'; c++) {
                        chars[l] = c;
                        String headDiff = String.valueOf(chars);
                        if (wordSet.contains(headDiff)){
                            if (headDiff.equals(endWord)) return layer + 1; // Find endword, return.
                            q.offer(headDiff);
                            wordSet.remove(headDiff);
                        }
                    }
                }
            }
        }

        return 0;
    }

}

ac3: bidirectional BFS

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        // edge cases
        if (wordList == null || wordList.size() == 0) return 0;

        Set<String> wordSet = new HashSet<>(wordList);
        if (!wordSet.contains(endWord)) return 0;

        Set<String> visited = new HashSet<>();
        Set<String> beginSet = new HashSet<>();
        beginSet.add(beginWord);
        Set<String> endSet = new HashSet<>();
        endSet.add(endWord);

        int len = 0;
        while (!beginSet.isEmpty() && !endSet.isEmpty()) {
            if (beginSet.size() > endSet.size()) {
                Set<String> tmp = beginSet;
                beginSet = endSet;
                endSet = tmp;
            }

            len++;
            Set<String> tmpSet = new HashSet<>();
            for(String w : beginSet) {
                char[] chars = w.toCharArray();
                for (int i = 0; i < chars.length; i++) {
                    char old = chars[i];
                    for (char c = 'a'; c <= 'z'; c++) {
                        chars[i] = c;
                        String newWord = String.valueOf(chars);
                        if (wordSet.contains(newWord)) {
                            if (endSet.contains(newWord)) return len + 1;
                            if (!visited.contains(newWord)) {
                                tmpSet.add(newWord);
                                visited.add(newWord);
                            }
                        }
                    }
                    chars[i] = old;
                }
            }
            beginSet = tmpSet;
        }


        return 0; // can't find
    }
}
/*
BFS
Queue<String> q
transform word 1 letter a step, lookup in dict, if contains, add to queue, if == endword return
return 0;
*/