<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 trie_node{
public:
    trie_node* next[26];
    int id;
    trie_node(){
        for(int i=0; i<26; i++) next[i] = nullptr;
        id = -1;
    }

    void insert(const string& word, int ID){
        trie_node* node = this;
        for(int i=0; i<word.size(); i++){
            int cur = word[i]-'a';
            if(!node->next[cur]) node->next[cur] = new trie_node();
            node = node->next[cur];
        }
        node->id = ID;
    }

    int find_id(const string& word){
        trie_node* node = this;
        for(int i=0; i<word.size(); i++){
            int cur = word[i]-'a';
            if(!node->next[cur]) return -1;

            node = node->next[cur];
        }
        return node->id;
    }
};

class Solution {
    static const bool cmp(string& a, string& b){
        if(a.size() != b.size()) return a.size()<b.size();
        return a<b;
    }
public:
    int longestStrChain(vector<string>& words) {
        int n = words.size();
        int max_len = 0;
        trie_node* root = new trie_node();
        sort(words.begin(), words.end(), cmp);
        vector<int> dp(n, 0);
        for(int i=0; i<words.size(); i++){
            int size = words[i].size();
            for(int remove=0; remove<size; remove++){
                string cur = words[i].substr(0, remove) + words[i].substr(remove+1);
                int id = root->find_id(cur);
                if(id==-1) continue;

                dp[i] = max(dp[i], dp[id]);
            }
            dp[i]++;
            max_len = max(max_len, dp[i]);
            root->insert(words[i], i);
        }

        return max_len;
    }
};
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