<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:
    string longestWord(vector<string>& words) {

        int max_len = 0;
        string res;
        queue<string> q;
        unordered_set<string> visited;
        unordered_set<string> all_words(words.begin(), words.end());

        for(auto s: words){
            if(s.length() == 1){
                q.push(s);
                visited.insert(s);
            }
        }
        cout<<max_len<<endl;
        while(!q.empty()){
            string curr = q.front();
            q.pop();
            cout<<curr<<endl;
            if(curr.length() >= max_len){
                if(curr.length() > max_len){
                    max_len = curr.length();
                    res = curr;
                }
                else if(curr.length() == max_len){
                    res = curr < res ? curr : res;
                }
                
            }
            for(char c = 'a'; c<='z'; c++){
                string new_word = curr + c;
                // cout<<new_word<<endl;
                if(all_words.count(new_word)){
                    q.push(new_word);
                    visited.insert(new_word);
                }
            }
        }

        return res;
        
    }
};
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