<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:
    bool canPick(string a, string b) {

        int diff = 0;
        for (int i = 0; i < a.length(); i++) {
            if(a[i] != b[i])
                diff += 1;
            
            if(diff > 1)
                return false;
        }

        return true;
    }
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        
        bool beginFound = false, endFound = false;

        for (auto it : wordList) {

            if (it == beginWord && beginFound != true)
                beginFound = true;
            if (it == endWord && endFound != true)
                endFound = true;

            if(beginFound && endFound)
                break;
        }

        if(!endFound)
            return 0;
        if(!beginFound)
            wordList.push_back(beginWord);
        
        int n = wordList.size();
        unordered_map<string, vector<string>> mp;

        for (int i = 0; i < n; i++) {
            for (int j = i+1; j < n; j++) {

                string u = wordList[i];
                string v = wordList[j];

                if(canPick(u, v)) {

                    mp[u].push_back(v);
                    mp[v].push_back(u);
                }
            }
        }

        queue<pair<string, int>> q;
        unordered_map<string, bool> vis;

        q.push({beginWord, 1});
        vis[beginWord] = true;

        while(!q.empty()) {

            string word = q.front().first;
            int dist = q.front().second;

            q.pop();

            auto vec = mp[word];

            for (auto it : vec) {

                if(vis.find(it) == vis.end()) {
                    vis[it] = true;
                    if(it == endWord)
                        return dist + 1;
                    q.push({it, dist+1});
                }
            }
        }

        return 0;
    }
};
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