<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 isPredecessor(string a, string b){
        int differences = 0;
        int n = a.size();
        int m = b.size();
        if(m != n+1){
            return false;
        }
        int ai = 0;
        int bi = 0;
        while(ai < n && bi < m){
            if(a[ai] != b[bi]){
                differences++;
                bi++;
                if(differences > 1){
                    return false;
                }
                // if(ai < n-1 && a[ai+1] == b[bi]){
                //     ++ai;
                // } else if(bi < m-1 && b[bi+1] == a[ai]){
                //     ++bi;
                // }
            } else{
                ai++;
                bi++;
            }
        }
        // cout<<b<<endl;
        if(bi == m-1 && ai == n){
            // cout<<b<<" good1"<<endl;
            return true;
        }
        if(differences == 1){
            // cout<<b<<" good2"<<endl;
            return true;
        }
        // cout<<b<<" again"<<endl;
        return false;
    }
    int longestStrChain(vector<string>& words) {

        sort(words.begin(), words.end(), [](string& a, string& b){
            return a.size() < b.size();
        });
        for(string s: words){
            cout<<s<<" ";
        }
        cout<<endl;
        const int n = words.size();
        int dp[1000] = {0};
        dp[0] = 1;
        // memset(dp, 1, 1000);
        // cout<<dp[1]<<endl;


        int maxlen = 1;
        for(int i = 0; i < n; ++i){
            if(dp[i] == 0){
                dp[i] = 1;
            }
            for(int j = i+1; j < n; ++j){
                if(words[j].size() - words[i].size() > 1){
                    break;
                }
                if(isPredecessor(words[i], words[j])){
                    // cout<<i<<" "<<j<<endl;
                    dp[j] = max(dp[j], dp[i] + 1);
                }
            }
            maxlen = max(maxlen, dp[i]);
        }
        for(int k = 0; k < n; ++k){
            cout<<dp[k]<<" ";
        }
        cout<<endl;
        return maxlen; 
    }
};
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