<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 compare(string str1, string str2){
        if(str1.length() < str2.length())
        return 1;
        return 0;
    }
    int longestStrChain(vector<string>& words) {
        int n  = words.size();
        sort(words.begin() , words.end(), [](string& a , string& b){return a.length() <b .length();});
        vector<int> n_dp(n ,1);
        for(int i =0;i<n;i++)
        {
            for(int j=0;j<i;j++)
            {
                int n1 = words[i].length();
                int n2 = words[j].length();
                if(n1-n2 !=1)continue;
                vector<vector<int>> dp(n1, vector<int>(n2 , -1));
                if(words[i][0]==words[j][0])
                {
                    dp[0][0]=1;
                }
                else
                {   
                    dp[0][0] =0;        
                }
                for(int z=1;z<n2;z++)
                {
                    if(words[i][0]==words[j][z])
                    {
                        dp[0][z] = 1;
                    }
                    else
                    {
                        dp[0][z] = dp[0][z-1];
                    }
                }
                for(int z=1;z<n1;z++)
                {
                    if(words[i][z]==words[j][0])
                    {
                        dp[z][0] = 1;
                    }
                    else
                    {
                        dp[z][0] = dp[z-1][0]; 
                    }
                }
                for(int ind1=1;ind1<n1;ind1++)
                {
                    for(int ind2=1;ind2<n2;ind2++)
                    {   
                        if(words[i][ind1] == words[j][ind2])
                        {
                            dp[ind1][ind2] =  1 + dp[ind1-1][ind2 -1];
                        }
                        else
                        {
                            dp[ind1][ind2] =  max(dp[ind1-1][ind2] ,dp[ind1][ind2-1]);
                        }
                    }
                }
                if(n1 -  dp[n1-1][n2-1] ==1)
                {
                    if(n_dp[j]+1 > n_dp[i])
                    {
                        n_dp[i] = n_dp[j]+1;
                    }
                }
            }
        }
        int max = -1;
        for(int i=0;i<n;i++)
        {
            if(n_dp[i]> max)
            {
                max = n_dp[i];
            }
        }
        return max;
    }
};
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