<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>

string ans="";
bool comp(string a , string b){
     if (a.length() == b.length()) {
        return a < b;  // Lexicographical comparison
    }
    return a.length() < b.length(); 
}
class Node{
    public:
    Node * brr[26]={NULL};
    bool end[26]={false};
};

class Trie{
    public:
    Node *root;
    Trie(){
        root=new Node();
    }


    void insert(string a){
        // string t="";
        int n=a.size();
        // char last=a[n-1];
        Node *temp=root;
        int i=0;

        for(;i<n;i++){

            int k=a[i]-'a';
            if(temp->brr[k]!=NULL && temp->end[k]){
                temp=temp->brr[k];
            }
            else{break;}
        }

        if(i==n-1){
            temp->brr[a[n-1]-'a']=new Node();
            temp->end[a[n-1]-'a']=true;   
            if(ans.size()<a.size()){ans=a;}
        }
        
    }
};

class Solution {
public:

  
    string longestWord(vector<string>& arr) {
        sort(arr.begin(), arr.end(), comp);
        int n=arr.size();
        Trie *t=new Trie();
        ans="";

        for(int i=0;i<arr.size();i++){
            string s=arr[i];
            if(i+1<n && arr[i]==arr[i+1]){continue;}
            t->insert(s);

        }

        return ans;
    }
};
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