<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 f(int index1, int index2, string &str1, string &str2, vector<vector<int>> &dp) {
    if (index1 < 0 || index2 < 0) {
        return 0;
    }

    if (dp[index1][index2] != -1) {
        return dp[index1][index2];
    }

    if (str1[index1] == str2[index2]) {
        return dp[index1][index2] = 1 + f(index1 - 1, index2 - 1, str1, str2, dp);
    } else {
        return dp[index1][index2] = max(f(index1 - 1, index2, str1, str2, dp), f(index1, index2 - 1, str1, str2, dp));
    }
}

int longestStrChain(vector<string> &nums) {
    if (nums.empty()) return 0;

    vector<int> rank(nums.size(), 1);
    vector<int> parent(nums.size());

    for (int i = 0; i < nums.size(); i++) {
        parent[i] = i;
    }

    sort(nums.begin(), nums.end(), [](const string &a, const string &b) {
        return a.size() < b.size();
    });

    for (int i = 1; i < nums.size(); i++) {
        for (int j = i - 1; j >= 0; j--) {
            if (nums[j].length() + 1 == nums[i].length()) {
                vector<vector<int>> dp(nums[i].size(), vector<int>(nums[j].size(), -1));
                if (f(nums[i].length() - 1, nums[j].length() - 1, nums[i], nums[j], dp) + 1 == nums[i].length()) {
                    if (rank[j] + 1 > rank[i]) {
                        rank[i] = rank[j] + 1;
                        parent[i] = j;
                    }
                }
            }
        }
    }

    return *max_element(rank.begin(), rank.end());
}
    
    
};
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