<p>Given an array of strings <code>words</code> representing an English Dictionary, return <em>the longest word in</em> <code>words</code> <em>that can be built one character at a time by other words in</em> <code>words</code>.</p>

<p>If there is more than one possible answer, return the longest word with the smallest lexicographical order. If there is no answer, return the empty string.</p>

<p>Note that the word should be built from left to right with each additional character being added to the end of a previous word.&nbsp;</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;w&quot;,&quot;wo&quot;,&quot;wor&quot;,&quot;worl&quot;,&quot;world&quot;]
<strong>Output:</strong> &quot;world&quot;
<strong>Explanation:</strong> The word &quot;world&quot; can be built one character at a time by &quot;w&quot;, &quot;wo&quot;, &quot;wor&quot;, and &quot;worl&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;a&quot;,&quot;banana&quot;,&quot;app&quot;,&quot;appl&quot;,&quot;ap&quot;,&quot;apply&quot;,&quot;apple&quot;]
<strong>Output:</strong> &quot;apple&quot;
<strong>Explanation:</strong> Both &quot;apply&quot; and &quot;apple&quot; can be built from other words in the dictionary. However, &quot;apple&quot; is lexicographically smaller than &quot;apply&quot;.
</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;= 30</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
</ul>

struct Node{
    map<char, Node*> ch;
    bool end=false;
};
class Trie{
    public:
    Node* root;
    Trie(){
        root= new Node();
        root->end = true;
    }
    void insert(string s){
        Node* temp= root;
        for(auto c: s){
            if(temp->ch.count(c)==0) temp->ch[c]= new Node();
            temp= temp->ch[c];
        }
        temp->end = true;
    }
    void call(Node* temp, string cur, string &ans){
        if(temp->end==false) return;
        if(cur.size()==ans.size()) ans= min(ans, cur);
        if(cur.size()>ans.size()) ans= cur;

        for(auto x: temp->ch){
            call(x.second, cur+x.first, ans);
        }
    }
};

class Solution {
public:
    string longestWord(vector<string>& words) {
        Trie trie;
        for(auto w: words) trie.insert(w);

        string ans="";
        trie.call(trie.root, "", ans);
        return ans;
    }
};
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