<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 Node{
    public:
    char data;
    bool isEnd;
    map<char,Node*>mp;
    Node(char data,bool isEnd){
        this->data=data;
        this->isEnd=isEnd;
    }
};
class Trie{
    public:
    Node* root;
    Trie(){
        root= new Node('\0',false);
    }
    void add(string s){
        if(s.size()==0) return;
        Node* t=root;
        for(int i=0;i<s.size();i++){
            if(t->mp.find(s[i])!=t->mp.end()){
                //the child node exixts already
                if(i==s.size()-1){
                    t->mp[s[i]]->isEnd=true;
                }
                t=t->mp[s[i]];
                continue;
            }else{
                //the child node did not exist so what we can do is build one
                Node* n;
                if(i==s.size()-1){
                    n=new Node(s[i],true);
                }else{
                    n=new Node(s[i],false);
                }
                t->mp[s[i]]=n;
                t=t->mp[s[i]];//movving to the next node
            }
        }
    }
    // pair<int,string> longestPath(){

    // }

};
class Solution {
public:
    string longestWord(vector<string>& words) {
        // Trie* t=new Trie();
        // for(auto x:words){
        //     t->add(x);
        // }
        // // now all we need to do is to return the longest path in this
        // pair<int,string>ans= t->longestPath();
        // // return ans.second;
        sort(words.begin(), words.end());
        set<string>st;
        string ans="";
        for(auto x:words) st.insert(x);
        for(auto x:words){
            bool isans=true;
        for (int i = 1; i < x.size(); i++) {
            if (st.find(x.substr(0, i)) == st.end()) {
                isans = false;
                break; // Exit the loop as soon as a prefix is missing
            }
        }
            if(isans&&x.size()>ans.size()||x.size()==ans.size()&&x<ans)ans=x;
        }
        return ans;
    }
};
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