<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>

class Solution {
public:
    class TrieNode {
    public:
	    TrieNode* children[26];
	    bool isEndOfWord;
	    TrieNode() {
		    for (int i = 0; i < 26; ++i) children[i] = nullptr;
		    isEndOfWord = false;
	    }
    };
class Trie {
private:
	TrieNode* root;
	
    void makesol(TrieNode* root,string& sol,string tmp,bool isFirst)
    {
        if (!root) return;
        if (!isFirst&&!root->isEndOfWord)
        return;
        for (int i=0;i<26;i++)
        {
            char s='a'+i;
            makesol(root->children[i],sol,tmp+s,0);
        }
        if (tmp.length()>sol.length())
        sol=tmp;
    }

public:
	Trie() {
		root = new TrieNode();
	}
	// Insert a word into the Trie
	void insert(const string& word) {
		TrieNode* node = root;
		for (char c : word) {
			int index = c - 'a';
			if (!node->children[index]) {
				node->children[index] = new TrieNode();
			}
			node = node->children[index];
		}
		node->isEndOfWord = true;
	}
	// Search for a word in the Trie
	string sol()
    {
        string sol="";
        string tmp="";
        makesol(root,sol,tmp,1);
        return sol;
    }
};

    string longestWord(vector<string>& words) {
        Trie root;
        for (int i=0;i<words.size();i++)
        {
            root.insert(words[i]);
        }
        return root.sol();
    }
};
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