<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 static comparator(string a, string b) {
        return a.size() < b.size();
    }
    bool canBeChained(string &a, string &b) {
        if(a.size() - b.size() != 1) {
            return false;
        }
        stack<char> s;
        for(auto it : b) {
            s.push(it);
        }
        int flag = 0;
        int i = a.size() - 1;
        for(; i >= 0; i --) {
            if(s.empty()) {
                break;
            } else {
                if(s.top() != a[i]) {
                    if(flag) {
                        return false;
                    }
                    flag = 1;
                } else {
                    s.pop();
                }
            }
        }
        if(i == 1) {
            return true;
        }
        if(s.empty()) {
            return true;
        }
        return false;
    }
    int longestStrChain(vector<string>& words) {
        sort(words.begin(), words.end(), comparator);
        int n = words.size();
        vector<int> dp(n, 1);
        // path(n);
        // for(int i = 0; i < n; i++) {
        //     path[i] = i;
        // }
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < i; j++) {
                if(canBeChained(words[i], words[j]) && dp[i] < dp[j] + 1) {
                    dp[i] = dp[j] + 1;
                    // path[i] = j;
                }
            }
        }
        // int maximum = 0;
        // int k  = 0;
        // for(int i = 0; i < n; i++ ) {
        //     if(dp[i] > maximum) {
        //         maximum = dp[i];
        //         k = i;
        //     }
        // }
        // while(path[k] != k) {
        //     cout<<words[k]<<" ";
        //     k = path[k];
        // }
        // cout<<words[k];
        return *max_element(dp.begin(), dp.end());
    }
};
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