<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 Solution:
    def longestWord(self, words: List[str]) -> str:
        class Node:
            def __init__(self):
                self.links=[None]*26
                self.cw=0
                self.cp=0
            def containsKey(self,K):
                return self.links[ord(K)-ord('a')]!=None
            def putKey(self,K,newNode):
                self.links[ord(K)-ord('a')]=newNode
            def getNext(self,K):
                return self.links[ord(K)-ord('a')]
            def incPrefixCount(self):
                self.cp=self.cp+1
            def incWordCount(self):
                self.cw=self.cw+1
            def getWordCount(self):
                return self.cw
        def insert(w,root):
            node=root
            for i in range(len(w)):
                if node.containsKey(w[i])==False:
                    node.putKey(w[i],Node())
                node=node.getNext(w[i])
                node.incPrefixCount()
            node.incWordCount()
        def satisfy(w):
            node=root
            for i in range(len(w)):
                node=node.getNext(w[i])
                if node.cw<1:
                    return False
            return True
        root=Node()
        ans=""
        for i in range(len(words)):
            insert(words[i],root)
        for i in range(len(words)):
            if satisfy(words[i]) and (len(words[i])>len(ans) or (len(words[i])==len(ans) and words[i]<ans)):
                    ans=words[i]
        return ans
                    
        
        
18300