<p>A <strong>transformation sequence</strong> from word <code>beginWord</code> to word <code>endWord</code> using a dictionary <code>wordList</code> is a sequence of words <code>beginWord -&gt; s<sub>1</sub> -&gt; s<sub>2</sub> -&gt; ... -&gt; s<sub>k</sub></code> such that:</p>

<ul>
	<li>Every adjacent pair of words differs by a single letter.</li>
	<li>Every <code>s<sub>i</sub></code> for <code>1 &lt;= i &lt;= k</code> is in <code>wordList</code>. Note that <code>beginWord</code> does not need to be in <code>wordList</code>.</li>
	<li><code>s<sub>k</sub> == endWord</code></li>
</ul>

<p>Given two words, <code>beginWord</code> and <code>endWord</code>, and a dictionary <code>wordList</code>, return <em>the <strong>number of words</strong> in the <strong>shortest transformation sequence</strong> from</em> <code>beginWord</code> <em>to</em> <code>endWord</code><em>, or </em><code>0</code><em> if no such sequence exists.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> beginWord = &quot;hit&quot;, endWord = &quot;cog&quot;, wordList = [&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;lot&quot;,&quot;log&quot;,&quot;cog&quot;]
<strong>Output:</strong> 5
<strong>Explanation:</strong> One shortest transformation sequence is &quot;hit&quot; -&gt; &quot;hot&quot; -&gt; &quot;dot&quot; -&gt; &quot;dog&quot; -&gt; cog&quot;, which is 5 words long.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> beginWord = &quot;hit&quot;, endWord = &quot;cog&quot;, wordList = [&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;lot&quot;,&quot;log&quot;]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The endWord &quot;cog&quot; is not in wordList, therefore there is no valid transformation sequence.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= beginWord.length &lt;= 10</code></li>
	<li><code>endWord.length == beginWord.length</code></li>
	<li><code>1 &lt;= wordList.length &lt;= 5000</code></li>
	<li><code>wordList[i].length == beginWord.length</code></li>
	<li><code>beginWord</code>, <code>endWord</code>, and <code>wordList[i]</code> consist of lowercase English letters.</li>
	<li><code>beginWord != endWord</code></li>
	<li>All the words in <code>wordList</code> are <strong>unique</strong>.</li>
</ul>

class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        # TC of this part: O(n^2 * m) where n = len(wordList) and m is the length of a word
        # this is too slow in LC
        # adj = {w: [] for w in wordList}
        # adj[beginWord] = []
        # for word in adj:
        #     for w in wordList:
        #         if self.wordDistance(word, w):
        #             adj[word].append(w)


        # TC of solution: O(n^2 * m) 

        # TC of this part: O(n * m^2). This one works in LC
        # n = len(wordList) m = len(word)
        # each word has m patterns
        # total number of patterns = n*m
        # time to construct 1 pattern = O(m)
        # Hence, TC of this part: O(n * m^2)

        # paterns for word "hot": ["*ot", "h*t", "ho*"]
        # adj = {"*ot": ["hot", "dot", "lot"], ...}
        wordList.append(beginWord)
        wordToPatterns = collections.defaultdict(list)
        adj = collections.defaultdict(list)
        for word in wordList:
            for j in range(len(word)):
                pattern = word[:j] + "*" + word[j+1:]   # O(m)
                wordToPatterns[word].append(pattern)
                adj[pattern].append(word)
        
        visited = set()
        def bfs(): 
            # O(V + E), V+E = n^2 in worst case
            # Time in each iteration: O(m)
            # Hence, TC: O(n^2 * m)
            q = deque([beginWord])
            visited.add(beginWord)

            level = 0
            while q:
                level += 1
                for i in range(len(q)):
                    word = q.popleft()
                    print(word)
                    if word == endWord:
                        return level

                    # neighbours of w
                    for pattern in wordToPatterns[word]:
                        for nei in adj[pattern]:
                            if nei not in visited:
                                q.append(nei)
                                visited.add(nei)

            return 0


        return bfs()

            
    # def wordDistance(self, w1, w2):
    #     dist = 0
    #     for i in range(len(w1)):
    #         if w1[i] != w2[i]: dist += 1
    #         if dist > 1:
    #             return False

    #     return dist == 1
25862