<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:
    void dfs(int node,string &crr,string &ans,vector<vector<int>>&trie,vector<int>&is_terminal){
        if(crr.length() > ans.length()) ans = crr;
        for(int ch = 0 ; ch < 26 ; ch ++){
            int neighbour = trie[node][ch];
            if(neighbour != 0 && is_terminal[neighbour]){
                crr.push_back(char(ch + 'a'));
                dfs(neighbour,crr,ans,trie,is_terminal);
                crr.pop_back();
            }
        }
    }
    string longestWord(vector<string>& words) {
        vector<vector<int>>trie(1,vector<int>(26,0));
        vector<int>is_terminal(1,0);
        int N = 0;
        for(auto it : words){
            int node = 0;
            for(auto c : it){
                if(trie[node][c - 'a']){
                    node = trie[node][c - 'a'];
                }
                else{
                    trie.push_back(vector<int>(26));
                    is_terminal.push_back(0);
                    trie[node][c - 'a'] = ++N;
                    node = trie[node][c - 'a'];
                }
            }
            is_terminal[node] = 1;
        }
        string crr = "",ans = "";
        dfs(0,crr,ans,trie,is_terminal);

        return ans;
    }
};
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