<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:
    // Function to calculate the length of the Longest Common Subsequence (LCS)
      int lcs(string &s1, string &s2) {
        int n = s1.size();
        int m = s2.size();

        vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0)); // Create a DP table initialized with 0

        // Fill in the DP table to calculate the length of LCS
        for (int ind1 = 1; ind1 <= n; ind1++) {
            for (int ind2 = 1; ind2 <= m; ind2++) {
                if (s1[ind1 - 1] == s2[ind2 - 1])
                    dp[ind1][ind2] = 1 + dp[ind1 - 1][ind2 - 1]; // Characters match, increment LCS length
                else
                    dp[ind1][ind2] = max(dp[ind1 - 1][ind2], dp[ind1][ind2 - 1]); // Characters don't match, consider the maximum from left or above
            }
        }

        return dp[n][m]; // Return the length of the Longest Common Subsequence
    }

  // Helper function for longestStrChain
    int fun(vector<string>& words, int n, int index, int last, vector<vector<int>>& dp) {
        if (index == n) {
            return 0;
        }

        if (dp[index][last + 1] != -1) { // Adjust for -1 index
            return dp[index][last + 1];
        }

        int nontaken = fun(words, n, index + 1, last, dp);
        int taken = 0;
        if (last == -1 || (words[index].length() == words[last].length() + 1 && lcs(words[index], words[last]) == words[last].length())) {
            taken = 1 + fun(words, n, index + 1, index, dp);
        }

        return dp[index][last + 1] = max(taken, nontaken);
    }

    // Function to find the longest string chain
    int longestStrChain(vector<string>& words) {
        int n = words.size();
        sort(words.begin(), words.end(), [](const string& a, const string& b) {
            return a.size() < b.size();
        });
        vector<vector<int>> dp(n, vector<int>(n + 1, -1));
        return fun(words, n, 0, -1, dp);
    }

};





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