<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 {
    bool isOneLetterDiff(const string& from, const string& to) {
        if(from.size() < to.size() - 1 || from.size() > to.size() + 1) {
            return false;
        }
        int changes = 0;
        int shortest = min(from.size(), to.size());
        for(int i=0;i < shortest; i++) {
            if(from[i] != to[i]) {
                changes++;
            }
        }
        if(from.size() != to.size()) {
            changes++;
        }
        return changes < 2 && changes != 0;
    }

    vector<string> getNeighbors(const string& from, const vector<string>& dict) {
        vector<string> res;
        for(const auto & word: dict) {
            if(isOneLetterDiff(from, word)) {
                res.push_back(word);
            }
        }
        return res;
    }

public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_map<string, vector<string>> dict;
        dict[beginWord] = getNeighbors(beginWord, wordList);
        for(const auto & word: wordList) {
            dict[word] = getNeighbors(word, wordList);
        }
        unordered_set<string> visited;
        queue<pair<int, string>> bfs;
        bfs.push({1, beginWord});
        while(!bfs.empty()) {
            auto [depth, word] = bfs.front();
            bfs.pop();
            if(word == endWord) {
                return depth;
            }
            visited.insert(word);
            for(const auto& neigh: dict[word]) {
                if(!visited.contains(neigh)) {
                    bfs.push({depth + 1, neigh});
                }
            }
        }
        return 0;
    }
};
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