<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:
    bool check(string word1, string word2, int len1, int len2)
    {
        // if(len1 == 1)
        // {
        //     if(word1[0] != word2[0] && word1[0] != word2[1]) return false;
        //     else return true;
        // }
        // int i = 0; int j = 0;
        // while(i < len1)
        // {
        //     if(word1[i] != word2[j]) break;
        //     i++;
        //     j++;
        // }
        // if(i == len1 - 1)
        // {
        //     return true;
        // }
        // j++;
        // if(word1.substr(i) == word2.substr(j)) return true;
        // else return false;
        int count = 0;
            int k = 0;
            char c = word1[0];
            for(int i = 0; i<len2; i++){

                if(word2[i] == c){
                    c = word1[++k];
                }else{
                    count++;
                    if(count>1){
                        return false;
                    }
                }

            }
            return true;
    }
    int longestStrChain(vector<string>& words) {
        sort(words.begin(), words.end(), [](const auto& a, const auto& b){return a.length() < b.length();});
        // for(string word: words)
        // {
        //     cout << word << " ";
        // }
        // cout << endl;
        int n = words.size();
        int res = 1;
        vector<int> dp(n, 1);
        for(int i = n - 2; i >= 0; i--)
        {
            string word1 = words[i];
            int len1 = word1.length();
            for(int j = i + 1; j < n; j++)
            {
                string word2 = words[j];
                int len2 = word2.length();
                if(len2 - len1 == 0) continue;
                if(len2 - len1 > 1) break;
                if(check(word1, word2, len1, len2))
                {
                    dp[i] = max(dp[i], dp[j] + 1);
                }
            }
            res = max(res, dp[i]);
        }
        // for(int i: dp)
        // {
        //     cout << i << " ";
        // }
        return res;
    }
};
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