<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:
    queue<pair<int, int>> q;
    bool checkadj(string a, string b) {
        int c=0;
        for(int i=0; i<a.size(); i++) {
            if(a[i]!=b[i]) {
                c++;
                if(c>1) return false;
            }
        }
        return true;
    }
    void addadj(string a, unordered_map<string, vector<int>>& umap, int b) {
        string temp;
        int n=a.size();
        for(int i=0; i<n; i++) {
            temp = a.substr(0, i) + "*" + a.substr(i+1, n-i-1);
            // cout<<temp<<" ";
            umap[temp].push_back(b);
        }
        // cout<<endl;
    }
    void bfs(vector<int> adj[], vector<int>& path, int node, int end, int ct) {
        path[node]=min(ct, path[node]);
        int n=adj[node].size();

        for(int i=0; i<n; i++) {
            if(adj[node][i]==end) {
                if(path[end]>ct+1){
                    path[end] = ct+1;
                }
            }
            if(path[adj[node][i]]>ct+1)
                q.push({adj[node][i], ct+1});
        }
    }
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        int n=wordList.size();
        int end=n;
        vector<int> adj[n+1];
        vector<int> path(n+1, 5001);
        unordered_map<string, vector<int>> umap;
        // wordList.push_back(beginWord);

        addadj(beginWord, umap, n);

        for(int i=0; i<n; i++) {
            addadj(wordList[i], umap, i);
            if(wordList[i]==endWord)
                end = i;
        }
        if(end==n) return 0;
        vector<int> temp;

        for(auto it:umap) {
            temp = it.second;
            for(int i=0; i<temp.size(); i++) {
                // cout<<temp[i]<<" ";
                for(int j=i+1; j<temp.size(); j++) {
                    adj[temp[i]].push_back(temp[j]);
                    adj[temp[j]].push_back(temp[i]);
                }
            }
            // cout<<endl;
        }

        // for(int i=0; i<n; i++) 
        //     for(int j=1; j<n; j++) 
        //         if(checkadj(wordList[i], wordList[j])) {
        //             adj[i].push_back(j);
        //             adj[j].push_back(i);
        //         }
        
        q.push({n, 1});
        while(!q.empty()) {
            bfs(adj, path, q.front().first, end, q.front().second);
            q.pop();
        }
        if(path[end]>5000) return 0;
        return path[end];

    }
};
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