<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(object):
    def ladderLength(self, beginWord, endWord, wordList):
        """
        :type beginWord: str
        :type endWord: str
        :type wordList: List[str]
        :rtype: List[List[str]]
        """

        #create a trie
        
        #take all unique letters in the word list
        #for every letter in beginWord, try replacing with every letter in unique_leters
        unique_letters = set()
        for word in wordList:
            for c in word:
                unique_letters.add(c)

        #for every letter in beginWord, attempt to replace with unique_Letter
        #convert word list to a set for faster look up
        word_set = set(wordList)
        #build a dictinoary denoting what are possible lookups 
        word_graph = defaultdict(set)
        queue = deque()
        queue.append(beginWord)
        visited = set()
        while queue:
            currWord = queue.popleft()
            for i in range(len(currWord)):
                for letter in list(unique_letters):
                    if letter != currWord[i]:
                        curr = currWord[:i] + letter + currWord[i+1:]

                        if curr in word_set:
                            word_graph[currWord].add(curr)
                        if curr in word_set and curr not in visited:
                            queue.append(curr)
                            visited.add(curr)
        queue = deque([[beginWord,[beginWord]]])
        visited = set()
        solutions = []
        while queue:
            #print(queue)
            currWord, currPath = queue.popleft()
            if currWord not in visited:

                if currWord == endWord:
                    
                    heapq.heappush(solutions,(len(currPath),currPath))
                
                visited.add(currWord)
                neighbors = list(word_graph[currWord])
                for word in neighbors:
                    queue.append([word,currPath + [word]])
        
        return heapq.heappop(solutions)[0] if solutions else 0
                    
                    # else:
                        # print(curr, currWord)

        
        #print(word_graph)
22862