<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 lcs(string& a, string& b, int i, int j, vector<vector<int>>& dp){
        if(i==a.size()||j==b.size()){
            return 0;
        }
        if(dp[i][j]!=-1) return dp[i][j];
        if(a[i]==b[j]){
            return dp[i][j]=1+lcs(a,b,i+1,j+1,dp);
        }
        else{
            return dp[i][j]=max(lcs(a,b,i+1,j,dp), lcs(a,b,i,j+1,dp));
        }
    }
    bool ch(string& a, string& b){
        if(b.size()-a.size()!=1){
            return 0;
        }
        vector<vector<int>> dp(a.size()+2,vector<int>(b.size()+2,-1));
        int l = lcs(a,b,0,0,dp);
        if(l!=a.size()){
            return 0;
        }
        return 1;
    }
    int solve(vector<string>& wrds, int ind, int pre,vector<vector<int>>& dp){
        if(ind==wrds.size()){
            return 0;
        }
        if(dp[ind][pre+1]!=-1){
            return dp[ind][pre+1];
        }
        // cout<<"hi";
        if(pre==-1 || ch(wrds[pre],wrds[ind])){
            return dp[ind][pre+1]=max(1+solve(wrds,ind+1,ind,dp), solve(wrds,ind+1,pre,dp));
        }else{
            return dp[ind][pre+1]=solve(wrds,ind+1,pre,dp);
        }
    }
    static bool cmp(string &a, string&b){
        if(a.size()==b.size()){
            return a<b;
        }
        return a.size()<b.size();
    }
    int longestStrChain(vector<string>& wrds) {
        sort(wrds.begin(),wrds.end(),cmp);
        
        vector<vector<int>> dp(wrds.size()+4,vector<int>(wrds.size()+4,-1));
        return solve(wrds,0,-1,dp);
    }
};
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