<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 isPred(string& a, string& b){
        map<string, int> m;
        if(a.size() <= b.size() || a.size() -1 != b.size() ) return false;
        vector<vector<int>> dp(a.size()+1, vector<int>(b.size()+1, 0));
        for(int i = 1; i <= a.size(); i++){
            for(int j = 1; j <= b.size(); j++){
                if(a[i-1] == b[j-1]){
                    dp[i][j] = 1 + dp[i-1][j-1];
                }else{
                    dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
                }
            }
        }
        if(dp[a.size()][b.size()] == b.size() && b.size() == a.size()-1) return true;
        return false;

    }

    int f(int ind, int prev_ind, vector<string>& arr, vector<vector<int>>& dp){
        if(ind == arr.size()) return 0;
        int y = 0;
        if(dp[ind][prev_ind+1] != -1) return dp[ind][prev_ind+1];
        if(prev_ind == -1 || isPred(arr[ind] , arr[prev_ind])){
            y = 1 + f(ind+1, ind, arr, dp);
        }
        int x = f(ind+1, prev_ind , arr, dp);
        return dp[ind][prev_ind+1] = max(x, y);
    }
    static bool cmp(string& a, string& b){
        return b.size() > a.size();
    }
    int longestStrChain(vector<string>& words) {
        sort(words.begin(), words.end(), cmp);
        vector<vector<int>> dp(words.size()+1, vector<int>(words.size()+2, -1));
        return f(0, -1, words, dp);
    }
};
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