<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 longestStrChain(vector<string>& words) {
        unordered_map<string, int> m;
        int idx = 0;
        for (int i = 0; i < words.size(); i++) {
            if (m.find(words[i]) == m.end()) {
                m[words[i]] = idx;
                idx++;
            }
        }
        vector<vector<int>> adj(idx);
        for (int i = 0; i < words.size(); i++) {
            for (int j = 0; j <= words[i].length(); j++) {
                for (int c = 0; c < 26; c++) {
                    string tmp = words[i];
                    tmp.insert(tmp.begin() + j, static_cast<char>(c+'a'));
                    if (m.find(tmp) != m.end()) {
                        adj[m[words[i]]].push_back(m[tmp]);
                    }
                }
            }
        }
        vector<int> in(idx, 0);
        for (int i = 0; i < idx; i++) {
            for (int j = 0; j < adj[i].size(); j++) {
                in[adj[i][j]]++;
            }
        }
        vector<int> topo_order;
        queue<int> q;
        for (int i = 0; i < idx; i++) {
            if (!in[i]) {
                topo_order.push_back(i);
                q.push(i);
            }
        }
        while (!q.empty()) {
            int cur = q.front();
            q.pop();
            for (int i = 0; i < adj[cur].size(); i++) {
                in[adj[cur][i]]--;
                if (!in[adj[cur][i]]) {
                    topo_order.push_back(adj[cur][i]);
                    q.push(adj[cur][i]);
                }
            }
        }
        vector<int> dp(idx, 1);
        for (int i = 0; i < topo_order.size(); i++) {
            int cur = topo_order[i];
            for (int j = 0; j < adj[cur].size(); j++) {
                if (dp[adj[cur][j]] < dp[cur] + 1) {
                    dp[adj[cur][j]] = dp[cur] + 1;
                }
            }
        }
        int ans = 1;
        for (int i = 0; i < idx; i++) ans = max(ans, dp[i]);
        return ans;
    }
};
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