<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 solve(string s, unordered_set<string>& vis, unordered_map<string, int>& ans,
    unordered_map<string, vector<string>>& mp){
        int maxi = 0;
        vis.insert(s);
        //cout<<s<<" ";

        for(string& v: mp[s]){
            if(vis.find(v) == vis.end()){
                int num = solve(v, vis, ans, mp);
                ans[v] = num;
                maxi = max(maxi, num);
            }
            else{
                maxi = max(maxi, ans[v]);
            }
        }
        return 1 + maxi;
    }

    int longestStrChain(vector<string>& words) {
        unordered_map<string, vector<string>> mp;
        unordered_set<string> st;

        for(string& s: words) st.insert(s);

        for(string s: st){
            int n = s.size();
            for(int i=0;i<=n;i++){
                for(char j = 'a'; j <= 'z'; j++){
                    string p;
                    if(i == 0){
                        p.push_back(j);
                        p += s;
                        if(st.find(p) != st.end()) mp[s].push_back(p);
                        continue;
                    }
                    int ind = i;
                    p = s.substr(0, ind);
                    p.push_back(j);
                    p += s.substr(ind);
                    if(st.find(p) != st.end()) mp[s].push_back(p);
                }
            }
        }

        unordered_set<string> vis;
        unordered_map<string, int> ans;
        for(string s : st){
            if(vis.find(s) == vis.end()){
                //cout<<"|"<<s<<" -> ";
                ans[s] = solve(s, vis, ans, mp);
            }
        }
        
        int res = 0;
        for(auto it : ans){
            // cout<<it.first<<" "<<it.second<<endl;
            res = max(res, it.second);
        }

        return res;

        return 0;
    }
};
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