<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:
    vector<vector<int>> dp;
    bool check(const string& word1, const string& word2) {
        if (word2.length() != word1.length() + 1)
            return false;
        int i = 0, j = 0;
        bool foundDifference = false;
        while (i < word1.length() && j < word2.length()) {
            if (word1[i] == word2[j]) {
                i++;
            } else {
                if (foundDifference) {
                    return false;
                }
                foundDifference = true;
            }
            j++;
        }

        return true;
    }
    int rec(int level, int last, vector<string>& words) {
        // base case
        if (level == words.size()) {
            return 0;
        }
        // cache check
        if (dp[level][last] != -1)
            return dp[level][last];
        // compute
        int ans = 0;
        // check if i can/cannot take
        // if last == 1001
        if (last == 1001) {
            ans = rec(level + 1, level, words) + 1;
        } else if (check(words[last], words[level])) {
            ans = rec(level + 1, level, words) + 1;
        }
        ans = max(ans, rec(level + 1, last, words));
        // save and return
        return dp[level][last] = ans;
    }
    int longestStrChain(vector<string>& words) {
        sort(words.begin(), words.end(), [](string& a, string& b) {
            if (b.length() > a.length())
                return true;
            return false;
        });
        dp.assign(1002, vector<int>(1002, -1));
        int ans = rec(0, 1001, words);
        // for (int i = 0; i < words.size(); i++) {
        //     for (int j = 0; j < words.size(); j++) {
        //         cout << dp[i][j] << " ";
        //     }
        //     cout << endl;
        // }
        return ans;
    }
};
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