<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:
    vector<Node*>links ;
    bool flag ;

    Node() {
        links.resize(26, NULL) ;
        flag = false ;
    }

    Node *getNextNode(char ch) {
        return links[ch-'a'] ;
    }

    void setNextNode(char ch) {
        links[ch-'a'] = new Node() ;
    }

    bool isContainNode(char ch) {
        return (links[ch-'a'] != NULL) ;
    }

    bool isEnd() {
        return flag ;
    }

    void setEnd() {
        flag = true ;
    }
};

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

    void insertWord(string &word)
    {
        int n = word.length() ;
        Node *node = root ;

        for(int i=0 ; i<n ; i++)
        {
            if(!node->isContainNode(word[i])) {
                node->setNextNode(word[i]) ;
            }
            node = node->getNextNode(word[i]) ;
        }
        node->setEnd() ;
    }

    bool isAllPrefixesPresent(string &word)
    {
        int n = word.length() ;
        Node *node = root ;

        for(int i=0 ; i<n ; i++)
        {
            if(!node->isContainNode(word[i]) || !node->getNextNode(word[i])->isEnd()) {
                return false ;
            }

            node = node->getNextNode(word[i]) ;
        }
        return true ;
    }
};

class Solution {
public:
    // L => Average Length of all the strings

    string longestWord(vector<string>& words) {
        Trie trie ;

        for(auto word: words) {  // TC = O(N*L)
            trie.insertWord(word) ;
        }

        string ansStr = "" ;
        for(auto word: words)  // TC = O(N*L)
        {
            int n = word.length() ;
            int m = ansStr.length() ;

            if(trie.isAllPrefixesPresent(word)) 
            {
                if(n > m) {
                    ansStr = word ;
                }
                else if(n == m) {
                    ansStr = min(ansStr, word) ; 
                }
            }
        }

        return ansStr ;
    }
};
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