<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 {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set<string> exists;
        // We can do this because the list of words are unique.
        for(string w : wordList) {
            exists.insert(w);
        }

        // Since this problem can be modeled as a graph, where the word that can transform to each other has an edge, we will use that property.
        // We can compute the 1 character difference by simulating changing each character 'a' <= 'z', and see if it's in our word list.
        queue<string> q;
        q.push(beginWord);
        int dist = 1; // We always start with 1 because it starts at 1, and needs 1 additional to convert to any word.
        
        // To prevent cycles in our BFS, we need to mark the words as seen.
        unordered_set<string> seen;

        while (!q.empty()) {
            int sz = q.size();

            // As we process each word level by level, each level are all the words that previous word can transform to. So we want to try and transform
            // all those words, and add the ones we can transform further.
            for(int i = 0; i < sz; ++i) {
                auto curr = q.front(); q.pop();
                seen.insert(curr);
                if (curr == endWord) {
                    return dist;
                }

                // Now we've started with beginWord, try to see if we can transform it to any of the words in wordList by changing every character
                // from a -> z, and if we have it in our word list, we can add it to our queue.
                for(int j = 0; j < curr.size(); ++j) {
                    char orig = curr[j];
                    // Try every character from a to z.
                    for(char c = 'a'; c <= 'z'; ++c) {
                        curr[j] = c;
                        // Check if the modified word is in our wordList, and that we haven't used it.
                        if (exists.count(curr) && !seen.count(curr)) {
                            q.push(curr);
                        }
                    }
                    // Undo in place modification of word, and go to next character
                    curr[j] = orig;
                }
            }
            // After each word we process/can transform to, update our distance.
            dist++;
        }
        // If we never transformed to reach endWord, there's no valid answer.
        return 0;
    }
};
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