<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:

    def __init__(self):
        self.is_word_end = False
        self.children = {}

    def insert(self, word: str) -> None:
        if len(word) < 1:
            self.is_word_end = True
        else:
            if word[0] not in self.children:
                self.children[word[0]] = Trie()
            self.children[word[0]].insert(word[1:])

    def search(self, word: str) -> bool:
        if len(word) < 1:
            return self.is_word_end
        return word[0] in self.children and self.children[word[0]].search(word[1:])

    def startsWith(self, prefix: str) -> bool:
        if len(prefix) < 1:
            return self.is_word_end or len(self.children) > 0
        return prefix[0] in self.children and self.children[prefix[0]].startsWith(prefix[1:])

class Solution:
    def longestWord(self, words: List[str]) -> str:
        self.trie = Trie()
        for word in words:
            self.trie.insert(word)
        
        longest_first = sorted(words, key = lambda word: (-len(word), word))
        print(longest_first)
        for word in longest_first:
            match = True
            for i in range(1, len(word)):
                if not self.trie.search(word[:i]):
                    match = False
                    break
            if match:
                return word
        return ""

18000