<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 TrieNode{
    public: 
    char data;
    TrieNode* children[26];
    bool isTerminal;

    TrieNode(char ch)
    {
        data=ch;
        for(int i=0;i<26;i++)
        children[i]=0;
        isTerminal=false;
    }
};

class Trie
{
    public:
    TrieNode* root;
    Trie(){
        root= new TrieNode('\0');
    }
    void insertUtil(TrieNode* root,string word)
    {
        if(word.length()==0)
        {
            root->isTerminal=true;
            return;
        }

        int index=word[0]-'a';
        TrieNode* child;

        if(root->children[index]!=NULL)
        {
            child=root->children[index];
        }
        else
        {
            child=new TrieNode(word[0]);
            root->children[index]=child;
        }
        insertUtil(child,word.substr(1));
    }

    void insertWord(string word)
    {
        insertUtil(root,word);
    }
    string findLongestWord()
    {
        string currWord="";
        string longestWord="";
        dfs(root,currWord,longestWord);
        return longestWord;
    }

    void dfs(TrieNode* root,string& currWord,string& longestWord)
    {
        if(!root)return;

        if(root->isTerminal)
        {
            if(currWord.length()>longestWord.length() ||(currWord.length()==longestWord.length() && currWord<longestWord))
            longestWord=currWord;
        }

        for(int i=0;i<26;i++)
        {
            if(root->children[i]&& root->children[i]->isTerminal)
            {
                currWord.push_back('a'+i);
                dfs(root->children[i],currWord,longestWord);
                currWord.pop_back();
            }
        }
    }
};

class Solution {
public:
    string longestWord(vector<string>& words) {

        Trie* t=new Trie;
        for(int i=0;i<words.size();i++)
        {
            t->insertWord(words[i]);
        }
        
        return t->findLongestWord();
        
    }
};
48378