<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:
    TrieNode* next[26];
    bool end;
    char c;
    TrieNode(char _c){
        for(int i = 0; i < 26; ++i){
            next[i] = nullptr;
            
        }
        c = _c;
        end = false;
    }

    //returns reference to next
    TrieNode* getOrInsert(char c){
        int idx = c-'a';
        TrieNode* ret = next[idx];
        if(!ret){
            ret = new TrieNode(c);
            next[idx] = ret;
        }
        return ret;
    }

    //returns the longest valid suffix
    string getLongestSuffix(){
        if(!end){
            return "";
        }
        string longest = "";

        for(int i = 0; i < 26; ++i){
            if(next[i]){
                string ans = next[i]->getLongestSuffix();
                if(ans.length() > longest.length()){
                    longest = ans;
                }
            }
        }
        return c+longest;
    }
};

class Trie {
        public:
    TrieNode* root;


    Trie(){
       root = new TrieNode('a'); 
       root->end = true;
    }

    void insert(string word){
        //Insert into the trie
        TrieNode* curr = root;
        for(int i = 0; i < word.length(); ++i){
            TrieNode* next = curr->getOrInsert(word[i]);
            curr = next;
        }
        curr->end = true;
    }

};

class Solution {
public:
    string longestWord(vector<string>& words) {
        //Add words to trie
        Trie trie;
        for(int i = 0; i < words.size(); ++i){
            trie.insert(words[i]);
        }

        //Iterate the trie and find the highest depth where words all end
        TrieNode* curr = trie.root;
        string bestWord = "";
        return trie.root->getLongestSuffix().substr(1);

    }
};
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