<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 {
public:
    struct trie{
        trie* next[26];
        bool en;
        trie ()
        {
            for(int i=0; i<26; i++) next[i] = nullptr;
            en = false;
        }
    };
    
    void build(string s, trie *tr)
    {
         for(int i=0; i<s.length(); i++)
         {
             if(!tr->next[s[i] - 'a']) tr->next[s[i] - 'a']= new trie();
             tr = tr->next[s[i] - 'a'];
         }
            tr->en = true;
        return;
    }
    
    string longestWord(vector<string>& words) {
        
        trie * tr = new trie(), *tp, *tp1;
        string res = "", t="", ti;
        
        for(auto s: words) build(s, tr);
        
        queue<pair<trie*, string>> q;
        q.push({tr, ""});
        
        while(q.size())
        {
            tp1 = q.front().first, t= q.front().second; q.pop();
            
            if(t.length() > res.length() ) 
                res = t;
            else if (t.length() == res.length() )
            {
                if(t < res) res = t;
            }
            
            for(int i=0; i<26; i++)
            { 
                tp = tp1;
                if(tp->next[i]) 
                {
                    tp = tp->next[i];
                    if(tp->en)
                        ti = t, ti += i+'a', q.push({tp, ti});
                }
            }
        }
        return res;
    }
};
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