<p>You are given an array of <code>words</code> where each word consists of lowercase English letters.</p>

<p><code>word<sub>A</sub></code> is a <strong>predecessor</strong> of <code>word<sub>B</sub></code> if and only if we can insert <strong>exactly one</strong> letter anywhere in <code>word<sub>A</sub></code> <strong>without changing the order of the other characters</strong> to make it equal to <code>word<sub>B</sub></code>.</p>

<ul>
	<li>For example, <code>&quot;abc&quot;</code> is a <strong>predecessor</strong> of <code>&quot;ab<u>a</u>c&quot;</code>, while <code>&quot;cba&quot;</code> is not a <strong>predecessor</strong> of <code>&quot;bcad&quot;</code>.</li>
</ul>

<p>A <strong>word chain</strong><em> </em>is a sequence of words <code>[word<sub>1</sub>, word<sub>2</sub>, ..., word<sub>k</sub>]</code> with <code>k &gt;= 1</code>, where <code>word<sub>1</sub></code> is a <strong>predecessor</strong> of <code>word<sub>2</sub></code>, <code>word<sub>2</sub></code> is a <strong>predecessor</strong> of <code>word<sub>3</sub></code>, and so on. A single word is trivially a <strong>word chain</strong> with <code>k == 1</code>.</p>

<p>Return <em>the <strong>length</strong> of the <strong>longest possible word chain</strong> with words chosen from the given list of </em><code>words</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;a&quot;,&quot;b&quot;,&quot;ba&quot;,&quot;bca&quot;,&quot;bda&quot;,&quot;bdca&quot;]
<strong>Output:</strong> 4
<strong>Explanation</strong>: One of the longest word chains is [&quot;a&quot;,&quot;<u>b</u>a&quot;,&quot;b<u>d</u>a&quot;,&quot;bd<u>c</u>a&quot;].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;xbc&quot;,&quot;pcxbcf&quot;,&quot;xb&quot;,&quot;cxbc&quot;,&quot;pcxbc&quot;]
<strong>Output:</strong> 5
<strong>Explanation:</strong> All the words can be put in a word chain [&quot;xb&quot;, &quot;xb<u>c</u>&quot;, &quot;<u>c</u>xbc&quot;, &quot;<u>p</u>cxbc&quot;, &quot;pcxbc<u>f</u>&quot;].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;abcd&quot;,&quot;dbqca&quot;]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The trivial word chain [&quot;abcd&quot;] is one of the longest word chains.
[&quot;abcd&quot;,&quot;dbqca&quot;] is not a valid word chain because the ordering of the letters is changed.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 1000</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 16</code></li>
	<li><code>words[i]</code> only consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    int longestStrChain(vector<string>& words) {
        unordered_map<string,int> m;
        int cur=1;
        int n=words.size();
        for(int i=0;i<n;i++){
            m[words[i]]=cur;
            cur++;
        }
        int src=0;
        int dest=n+1;
        vector<int> adj[n+2];
        for(int i=1;i<=n;i++){
            adj[src].push_back(i);
            adj[i].push_back(dest);
        }
        for(int i=0;i<n;i++){
            string str=words[i];
            string cur="";
            for(int j=0;j<str.length();j++){
                string y=str.substr(j);
                for(char k='a';k<='z';k++){
                    cur.push_back(k);
                    string temp=cur;
                    string xy=temp.append(y);
                    if(m.count(xy)){
                        adj[i+1].push_back(m[xy]);
                    }
                    
                    cur.pop_back();
                }
                cur.push_back(str[j]);
            }
            for(char k='a';k<='z';k++){
                cur.push_back(k);
                if(m.count(cur)){
                    adj[i+1].push_back(m[cur]);
                }
                cur.pop_back();
            }
        }

        int dist[n+2];
        for(int i=0;i<=n+1;i++){
            dist[i]=-1e5;
        }
        dist[0]=0;
        priority_queue<pair<int,int>> pq;
        pq.push({0,0});
        while(pq.size()!=0){
            auto p=pq.top();
            pq.pop();
            int d=p.first;
            int node=p.second;
            if(dist[node]>d){
                continue;
            }
            for(int i=0;i<adj[node].size();i++){
                int nei=adj[node][i];
                if(dist[nei]<d+1){
                    dist[nei]=d+1;
                    pq.push({d+1,nei});
                }
            }
        }
        return dist[n+1]-1;
    }
};
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