<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] = NULL;
        isTerminal = false;
    }
};

class Trie
{
public:
    TrieNode *root;

    Trie()
    {
        root = new TrieNode('\0');
    }

    void insertUtil(TrieNode *root, string word)
    {
        // Base case
        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;
        }

        // Recursion
        insertUtil(child, word.substr(1));
    }

    void insertWord(string word)
    {
        insertUtil(root, word);
    }

    bool searchWord(string word) {
        TrieNode* curr = root;

        for (char ch : word) {
            int index = ch - 'a';
            if (curr->children[index] == NULL || !curr->children[index]->isTerminal) return false;
            curr = curr->children[index];
        }
        return true;
    }
};

class Solution {
public:
    string longestWord(vector<string>& words) {
        Trie* trie = new Trie();

        for(string& word : words) trie->insertWord(word);

        string longest = "";
        sort(words.begin(),words.end());
        
        for(string& word : words){
            if(trie->searchWord(word)){
                if (word.length() > longest.length()) longest = word;
            }
        }

        return longest;
    }
};
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