<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 longestStrChain(vector<string>& words) {
        
        int ans = 1;
        vector<vector<string>> wordss;
        wordss.resize(17);
        
        // word -> longest chain to that word
        unordered_map<string, int> mp;
        
        // Split up the words by their sizes
        for(string s : words){
            wordss[s.size()].push_back(s);
            mp[s] = 1;
        }
        
        // For each word in each size band
        for(int x = 0; x < wordss.size()-1; ++x){
            for(int y = 0; y < wordss[x].size(); ++y){
                
                //Check the size band above it
                for(int z = 0; z < wordss[x+1].size(); ++z){
                    
                    if(possible(wordss[x][y], wordss[x+1][z])){
                        mp[wordss[x+1][z]] = max(mp[wordss[x+1][z]], mp[wordss[x][y]]+1);
                        ans = max(mp[wordss[x+1][z]], ans);
                    }
                }
            }
            
        }
        
        return ans;
    }
    
    bool possible(string w1, string w2){

        int count = -1;
        for(int i = 0; i < w1.size(); ++i){
            if(w1[i] != w2[i]){
                count = i;
                break;
            }
        }
        
        if(count == -1)
            return true;
        
        for(int i = count; i < w1.size(); ++i){
            if(w1[i] != w2[i+1]){
                return false;
            }
        }
        
        return true;
    }
};
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