<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>

from collections import defaultdict, deque
class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        if beginWord == endWord:
            return 1
        def compareWords(word1, word2):
            difCount = 0 
            for i in range(len(word1)):
                if word1[i] != word2[i]:
                    difCount += 1
            if difCount > 1:
                return False 
            else:
                return True 
        
        if endWord not in wordList:
            return 0

        # 1. create a graph using wordList. n^2. space: n^2 
        wordConnections = defaultdict(list)
        for i in range(len(wordList) - 1):
            for j in range(i + 1, len(wordList)):
                # check if wordList[i], j is 1 char apart
                canConnect = compareWords(wordList[i], wordList[j])
                if canConnect:
                    wordConnections[wordList[i]].append(wordList[j])
                    wordConnections[wordList[j]].append(wordList[i])


        # 2. traverse it with beginWord:
        wordQueue = deque([]) # (currWord, pathCount)
        visitedWords = set()
        #a. find all words that are 1char different from beginWord
        for i in range(len(wordList)):
            if compareWords(beginWord, wordList[i]):
                wordQueue.append((wordList[i], 1))
      # Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
        while wordQueue:
                # 1, 4
            for i in range(len(wordQueue)):
                currWord, currPathCt = wordQueue.popleft()
                if currWord == endWord:
                    return currPathCt + 1
                visitedWords.add(currWord)

                nextWords = wordConnections[currWord]
                for nxtWord in nextWords:
                    if nxtWord not in visitedWords:
                        wordQueue.append((nxtWord, currPathCt + 1))
            # [word1, word2] --> word1, 
            # visitedWords = []
        return 0
19362