<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 helper(unordered_set<string>& words_set, string s, unordered_map<string, int>& dp){
        if(s.length() == 1){
           return 1; 
        }
        
        if(dp.count(s) != 0)
            return dp[s];
        
        int ans = 1;
        for(int i = 0; i < s.length(); i++){
            string temp = s.substr(0, i) + s.substr(i+1);
            if(words_set.count(temp) != 0){
                ans = max(ans, 1 + helper(words_set, temp, dp));
            }
        }
        dp[s] = ans;
        return ans;
        
    }
    
    int longestStrChain_recursion(vector<string>& words) {
        unordered_set<string> words_set;
        unordered_map<string, int> dp;
        for(string s: words){
            words_set.insert(s);
        }
        
        int ans = 0;
        for(string s: words){
            ans = max(ans, helper(words_set, s, dp));
        }
        
        return ans;
    }
    
    static bool compare(const string& s1, const string& s2){
        return (s1.size() < s2.size());
    }
    
    int longestStrChain(vector<string>& words) {
        unordered_set<string> uset(words.begin(), words.end());
        
        sort(words.begin(), words.end(), compare);
        
        unordered_map<string, int> dp;
        
        int ans = 0;
        
        for(int i = 0; i < words.size(); i++){
            string s = words[i];
            for(int j = 0; j < s.size(); j++){
                string temp  = s.substr(0, j) + s.substr(j+1);
                if(uset.count(temp) != 0){
                    dp[s] = max(dp[s], dp[temp] + 1);
                }
            }
            ans = max(ans, dp[s]);
        }
        
        return ans +1;
    }
    
    
};
















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