<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>

// create the comparator
bool comparator(string &str1,string &str2){
    return str1.size()<str2.size();
}

class Solution {
public:
    // function 
    bool checkconditions(string &str1,string &str2){
        // find the value of size1 and the value of size2
        int size1=str1.size();
        int size2=str2.size();

        // if the size2-size1 is less than 1
        if(size1-size2!=1){
            return false;
        }

        // create the varaible diff
        int diff=0;

        // create the variable indx1 and the variable indx2
        int indx1=0;
        int indx2=0;

        // while the value of indx is less than size1 and the value of indx2 is less than size2
        while(indx1<size1 && indx2<size2){
            // if the character matches
            if(str1[indx1]==str2[indx2]){
                indx1++;
                indx2++;
            }

            // if the charcter does not match
            else if(str1[indx1]!=str2[indx2]){
                diff++;
                indx1++;
            }

            // if the value of diff is more than 1
            if(diff>1){
                return false;
            }
        }

        // return true
        return true;
    }

    int longestStrChain(vector<string>& words) {
        // find the value of n
        int n=words.size();

        // create the vector dp
        vector<int> dp(n,1);

        // create the varaible maxlen
        int maxlen=INT_MIN;

        // sort the vector words
        sort(words.begin(),words.end(),comparator);

        // traverse the vector dp
        for(int i=0;i<n;i++){
            // find the string str1
            string str1=words[i];
            for(int j=0;j<i;j++){
                // find the string str2
                string str2=words[j];

                // if the value of the function call checkcondition(str1,str2) and the value of dp[j]+1 is more than dp[i]
                if(checkconditions(str1,str2) && dp[j]+1>dp[i]){
                    dp[i]=dp[j]+1;
                }
            }
            // update the value of maxlen
            maxlen=max(maxlen,dp[i]);
        }

        // return maxlen
        return maxlen;
    }
};
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