<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:
    class comp{
        public:
        bool operator()(string &a, string &b){
            return a.size() < b.size();
        }
    };

    bool compareStr(string &s1, string &s2){
        int i = 0;
        int j = 0;
        int n1 = s1.size();
        int n2 = s2.size();
        if(n2-n1 != 1){
            return false;
        }

        while(j < n2){
            if(s1[i] == s2[j]){
                i++;
                j++;
            }
            else{
                j++;
            }
        }
        if(i == n1 && j == n2){
            return true;
        }
        return false;
    }

    int solve(vector<string> &words, int i, int parent, vector<vector<int>> &dp){
        if(i == words.size()){
            return 0;
        }
        if(dp[i][parent+1] != -1){
            return dp[i][parent+1];
        }

        int incl = 0;
        int excl = 0;

        if(parent == -1){
            incl = 1 + solve(words, i+1, i, dp);
        }
        else if(compareStr(words[parent], words[i])){
            incl = 1 + solve(words, i+1, i, dp);
        }
        excl = 0 + solve(words, i+1, parent, dp);

        dp[i][parent+1] = max(incl , excl);
        return dp[i][parent+1];
    }

    int solveTab(vector<string> &words){
        int n = words.size();
        vector<vector<int>> dp(n+1 , vector<int>(n+2, 0));

        for(int i = n-1; i>=0; i--){
            for(int j = n-1; j >= -1; j--){
                int incl = 0;
                int excl = 0;

                if(j == -1){
                    incl = 1 + dp[i+1][i+1];
                }
                else if(compareStr(words[j], words[i])){
                    incl = 1 + dp[i+1][i+1];
                }
                excl = 0 + dp[i+1][j+1];

                dp[i][j+1] = max(incl , excl);
            }
        }
        return dp[0][0];
    }

    int longestStrChain(vector<string>& words) {
        int n = words.size();
        sort(words.begin(), words.end(), comp());
        vector<vector<int>> dp(n , vector<int>(n+1, -1));
        // return solve(words, 0, -1, dp);
        return solveTab(words);
    }
};
154834