<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:
    vector<Trie*> next;
    bool isEnd;
    Trie(){
        isEnd = false;
        next = vector<Trie*>(26);
    }

    void insert(vector<string> &words){
        for(auto word:words){
            Trie *current = this;
            for(int i = 0;i < word.size();i++){
                if(!current -> next[word[i] - 'a']){
                    current -> next[word[i] - 'a'] = new Trie;
                }
                current = current -> next[word[i] - 'a'];
                if(i == word.size() - 1)current->isEnd = true;
            }
        }
    }
    //return true for those words than can be built
    bool build(string searchWord){
        Trie* current = this;
        for(int i = 0;i < searchWord.size();i++){
            if(!current -> next[searchWord[i] - 'a']){
                return false;
            }
            current = current -> next[searchWord[i] - 'a'];
            if(!current -> isEnd)return false;
        }        
        return current -> isEnd;
    }
};

class GreatL{
    public:
    bool operator()(const string& a,const string& b)const{
        if(a.size() == b.size()){
            return a > b;
        }
        return a.size() < b.size();
    }
};

class Solution {
public:
    string longestWord(vector<string>& words) {
        priority_queue<string, vector<string>, GreatL> longString;
        sort(words.begin(),words.end());
        Trie *root;
        root = new Trie;
        root->insert(words);

        for(auto word:words){
            if(root->build(word)){
                longString.push(word);
            }
        }     
        if(!longString.size())return "";
        return longString.top();  
    }
};
42003