<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:
        
        
        word_ch = {}
        for w in words:
            word_ch[w] = len(w)
        
        new_word=sorted(word_ch.items(), key=lambda item: item[1])
        
        ch = {}
     
        for w1 in new_word:
            w = w1[0]
            if len(w) == 1:
                ch[w] = []
            else:
                tmp = w[:-1] 
                if tmp in ch:
                    ch[tmp].append(w)
                    ch[w] = []
                    
                    
        max_word = []
        max_len = 0
        
        def dfs(k):
            
            nonlocal ch
            nonlocal max_len
            nonlocal max_word
            
            if len(ch[k]) == 0:
                return k
            
            else:
                
                for itm in ch[k]:
                    w12 = dfs(itm)
                    print("w12",w12)
                    if w12:
                        if len(w12) > max_len:
                            max_word = []
                            max_word.append(w12)
                            max_len = len(w12)
                        elif len(w12) == max_len and w12 != max_word[0]:
                            max_word.append(w12)
                            
                return w12
                         
        for k,v in ch.items():
            
            if len(k) == 1:
                
                tp = dfs(k)
                if not max_word:
                    max_word.append(tp)
                
        
        
        if max_word:
            return sorted(max_word)[0]
        else:
            return ''
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