<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:
        if endWord not in wordList:
            return 0

        wordMap = defaultdict(list)
        # or just visited to denote both words and groups 
        visited = set()
        for word in wordList:
            word_arr = [s for s in word]
            for i in range(len(word_arr)):
                wordMap[''.join(word_arr[:i] + ['*'] + word_arr[i+1:])].append(word)
        
        start = [s for s in beginWord]
        queue = deque()
        for i in range(len(start)):
            queue.append((''.join(start[:i] + ['*'] + start[i+1:]), 1))

        while queue:
            # word represents *ot, h*t, etc
            word, length = queue.popleft()
            # transformed represents hot, dot, lot (the values of wordMap)
            for transformed in wordMap[word]:
                if transformed == endWord:
                    return length + 1
                if transformed not in visited:
                    visited.add(transformed)
                    new_word = [s for s in transformed]
                    for i in range(len(new_word)):
                        new_transformed = ''.join(new_word[:i] + ['*'] + new_word[i+1:])
                        if new_transformed not in visited:
                            visited.add(new_transformed)
                            queue.append((new_transformed, length + 1))
        return 0 

        # "*it", "h*t", "hi*" in queue 
        # "h*t" -> "hot" (1)
        # "hot" -> "*ot", "h*t", "ho*" (We've seen h*t) 
        # "*ot" -> "hot", "dot", "lot" (we've seen hot) (2)
        # "dot", "lot" -> "*ot", "d*t", "do*", "*ot", "l*t", "lo*" 
        # "do*", "l*t", "lo*" -> "dot", "dog", "lot" (we've seen dot, lot) (3)
        # "dog" -> "*og", "d*g", "do*" -> (we've not seen any) ()
        # "*og", "d*g", "do*" -> "dog", "log", "cog" (found!)
        
        queue = deque()
        return 1



        

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