<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 Trie {
public:
    bool isWord;
    vector<Trie*> children;
    Trie(){
        isWord = false;
        for(int i = 0;i < 26; i++){
            children.push_back(NULL);
        }
    }
};

class Solution {
public:
    int max_len = 0;
    string res = "";
    void DFS(Trie* root, int len, string curr){
        if(root->isWord){
            if(max_len < len){
                max_len = len;
                res = curr;
            }else if (len == max_len && curr < res) {
                res = curr;
            }
        }

        for(int i = 0; i < 26; i++){
            if(root->children[i] != NULL  && root->children[i]->isWord){
                string curr_char(1, i + 'a');
                DFS(root->children[i], len + 1, curr + curr_char);
            }
        }
    }
    string longestWord(vector<string>& words) {
        Trie* root = new Trie();
        for(int i = 0; i < words.size(); i++){
            Trie* curr = root;
            for(int j = 0; j < words[i].length(); j++){
                if(curr->children[words[i][j] - 'a'] == NULL){
                    curr->children[words[i][j] - 'a'] = new Trie();
                }
                curr = curr->children[words[i][j] - 'a'];
            }
            curr->isWord = true;
        }


        for(int i = 0; i < 26; i++){
            if(root->children[i] != NULL && root->children[i]->isWord){
                int word_count = 0;
                string curr(1, i + 'a');
                DFS(root->children[i], 1, curr);
            }
        }
        return res;
    }
};
57409