<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 lcs(string &s, string &t)
    {
        int n = s.size();
        int m = t.size();
        vector<int> dp(m + 1);
        vector<int> temp(m + 1);

        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {
                if (s[i - 1] == t[j - 1])
                    temp[j] = 1 + dp[j - 1];
                else
                    temp[j] = max(dp[j], temp[j - 1]);
            }
            dp = temp;
        }

        return dp[m];
    }

    bool isvalidstring(string &s, string &t)
    {

        if (s.length() == t.length())
            return false;

        int n = s.length();
        int m = t.length();

        int maxmatch = lcs(s, t);

        if ((n + m) - 2 * maxmatch == 1)
            return true;
        return false;
    }

    bool static cmp(string &s, string &t)
    {
        return s.length() < t.length();
    }

    int longestStrChain(vector<string> &words)
    {
        sort(words.begin(), words.end(), cmp);
        int n = words.size();
        vector<vector<int>> dp(n + 1, vector<int>(n + 1));

        // return findways(words, 0, -1, dp);

        for (int i = 0; i <= n; i++)
            dp[n][i] = 0;

        for (int idx = n - 1; idx >= 0; idx--)
        {
            for (int previdx = idx - 1; previdx >= -1; previdx--)
            {
                if (previdx < 0 || isvalidstring(words[previdx], words[idx]))
                {
                    int take = 1 + dp[idx + 1][idx + 1];
                    int nottake = dp[idx + 1][previdx + 1];
                    dp[idx][previdx + 1] = max(take, nottake);
                }
                else
                    dp[idx][previdx + 1] = dp[idx + 1][previdx + 1];
            }
        }

        return dp[0][0];
    }
};

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