<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 isAdjacent(string a, string b) {
        int n = a.length();
        int diff=0;
        for(int i=0;i<n;i++) diff += (a[i] != b[i] ? 1: 0 );
        if(diff==1) return true;
        return false;
    }


    int shortestDistance(int u, int v, vector <vector <int>>& gra, int n) {
        int d[n+1];
        for(int i=0;i<n;i++) d[i] = INT_MAX;
        d[u] = 0;
        int vis[n+1];
        for(int i=0;i<n;i++) vis[i]=0;
        
        set <pair<int,int>> s;
        for(int i=0;i<n;i++) s.insert({d[i], i});

        while(!s.empty()) {
            auto pp = *s.begin();
            s.erase(pp);
            int node = pp.second;
            int val = pp.first;
            if(val == INT_MAX) break;
            vis[node]=1;
            for(auto x: gra[node]) {
                if(vis[x]) continue;
                if(val + 1 < d[x] ) {
                    s.erase({d[x], x});
                    d[x] = val + 1;
                    s.insert({d[x], x});
                }
            }
        }
        if(d[v] == INT_MAX) return 0;
        else return d[v]+1;
    }


    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        int n = wordList.size();
        unordered_map <string, int> ind;
        for(int i=0;i<n;i++) ind[wordList[i]] = i;

        int found=0;
        for(int i=0;i<n;i++) {
            if(wordList[i] == beginWord) {found=1;break;}
        }
        if(!found) {
            wordList.push_back(beginWord);
            ind[beginWord] = n;
            n++;
        }

        found=0;
        for(int i=0;i<n;i++) {
            if(wordList[i] == endWord) {found=1;break;}
        }
        if(!found) {
           return 0;
        }

        vector <vector <int>> gra(5003);
        for(int i=0;i<n-1;i++) {
            for(int j=i+1;j<n;j++) {
                if(isAdjacent(wordList[i], wordList[j])) {
                    gra[i].push_back(j);
                    gra[j].push_back(i);
                }
            }
        }
        return shortestDistance(ind[beginWord], ind[endWord], gra,n);

    }
};
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