<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:
        """
        Brute force a Dijkstra's approach for now.
        As expected, gets TLE on Leetcode.
        """
        # Construct the adjacency list.
        wordSet = set(wordList)
        wordSet.add(beginWord)
        if endWord not in wordSet:
            return 0
        adj_list = self.adjacency_list(beginWord, endWord, wordList)
        # print(adj_list)

        distances = dict()
        for word in wordSet:
            distances[word] = float('inf')
        distances[beginWord] = 1

        queue = [(1, beginWord)]

        while queue:
            curr_dist, curr_word = heapq.heappop(queue)

            # If a shorter path to the current node has been found, don't process it.
            if curr_dist > distances[curr_word]:
                continue

            # Otherwise, iterate over each neighbor of the current node.
            for neighbor, weight in adj_list[curr_word].items():
                dist = curr_dist + weight

                # Only consider path if it's better than any path previously found.
                if dist < distances[neighbor]:
                    distances[neighbor] = dist
                    heapq.heappush(queue, (dist, neighbor))

        out = distances[endWord]
        return out if out != float('inf') else 0
    
    def adjacency_list(self, beginWord, endWord, wordList):
        adj_list = defaultdict(dict)
        wordList = [beginWord] + wordList
        # Could definitely be optimized by checking only letters available
        # in each index.
        # wordSet.add(beginWord)
        # wordSet.add(endWord)
        for i in range(len(wordList)):
            for j in range(i + 1, len(wordList)):
                word = wordList[i]
                word2 = wordList[j]
                if self.wordDist(word, word2) == 1:
                    adj_list[word][word2] = adj_list[word2][word] = 1
        return adj_list

    def wordDist(self, word1, word2):
        # Returns the distance between two words.
        total = 0
        for i in range(len(word1)):
            if word1[i] != word2[i]:
                total += 1
        return total
        
18487