<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 re(const unordered_map<string, vector<string>>& adjList, const string& start, const string& end) {
        if (start == end) return 0;

        unordered_map<string, int> distances;
        for (const auto& pair : adjList) {
            distances[pair.first] = INT_MAX ;
        }

        queue<string> q;
        q.push(start);
        distances[start] = 0;

        while (!q.empty()) {
            const string current = q.front();
            q.pop();

            for (const string& neighbor : adjList.at(current)) {
                if (distances[neighbor] == numeric_limits<int>::max()) {
                    distances[neighbor] = distances[current] + 1;
                    q.push(neighbor);

                    if (neighbor == end) {
                        return distances[neighbor];
                    }
                }
            }
        }

        return -1;
    }

    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_map<string, vector<string>> adjList;
        wordList.push_back(beginWord);
        bool endWordInList = false;

        for (const auto& word : wordList) {
            if (word == endWord) {
                endWordInList = true;
            }
            adjList[word] = {};
            for (const auto& potentialNeighbor : wordList) {
                if (isOneLetterDifferent(word, potentialNeighbor)) {
                    adjList[word].push_back(potentialNeighbor);
                    adjList[potentialNeighbor].push_back(word);
                }
            }
        }

        if (!endWordInList) {
            return 0;
        }

        int distance = re(adjList, beginWord, endWord);
        return distance != 0 ? distance + 1 : 0;
    }

private:
    bool isOneLetterDifferent(const string& word1, const string& word2) {
        int diffCount = 0;
        for (size_t i = 0; i < word1.size(); ++i) {
            if (word1[i] != word2[i]) {
                ++diffCount;
                if (diffCount >= 2) {
                    return false;
                }
            }
        }
        return diffCount == 1;
    }
};

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