<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:
    bool isEndOfWord = false;
    vector<TrieNode*> children;
    TrieNode(): children(26, nullptr) {
    }
};

class Trie {
public:
    TrieNode* root = new TrieNode();

    Trie() {
        
    }
    
    void insert(string word) {
        int i = 0; 
        TrieNode* curr = root;

        while(i < word.size()) {
            if(curr->children[word[i] - 'a'] == NULL) 
                curr->children[word[i] - 'a'] = new TrieNode();
            curr = curr->children[word[i] - 'a'];
            i++;
        }
        curr->isEndOfWord = true;
    }
    
    bool search(string word) {
        int i = 0; 
        TrieNode* curr = root;

        while(i < word.size()) {
            if(curr->children[word[i] - 'a'] == NULL) 
                return false;
            curr = curr->children[word[i] - 'a'];
            i++;
        }
        return curr->isEndOfWord == true;
    }
    
    bool startsWith(string prefix) {
        int i = 0; 
        TrieNode* curr = root;

        while(i < prefix.size()) {
            if(curr->children[prefix[i] - 'a'] == NULL) 
                return false;
            curr = curr->children[prefix[i] - 'a'];
            i++;
        }
        return true;
    }
};

class Solution {
public:
    string getLongestWord(TrieNode* root) {
        if(!root) return "";
        
        vector<TrieNode*> children = root->children;
        int maxLen = 0;
        string ans = "";
        
        for(int i = 0; i<26; i++) {
            if(children[i] && children[i]->isEndOfWord) {
                string temp = getLongestWord(children[i]);
                string x = "";
                x += ('a' + i);
                temp = x + temp;
                if(maxLen < temp.size()) {
                    maxLen = temp.size();
                    cout<<temp<<endl;
                    ans = temp;
                }
            }
        }
        return ans;
    }

    string longestWord(vector<string>& words) {
        Trie dictionary;
        for(auto word: words) dictionary.insert(word);

        return getLongestWord(dictionary.root);
    }
};
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