<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) {
        unordered_map<string, set<string>> adj;
        vector<string> singWord;
        int n= words.size();
        for(int i=0; i<n; i++){
            if(words[i].size() == 1){
                singWord.push_back(words[i]);
            }
            for(int j=0; j<n; j++){
                if(check(words[i], words[j])){
                    adj[words[i]].insert(words[j]);
                }
            }
        }
        int maxLen = 0;
        for(auto it: singWord){
            int currLen = 1;
           dfs(it, adj, currLen);
        }
        return ans;
    }

    string ans = ""; int maxLen = 0;
    void dfs(string u, unordered_map<string, set<string>> &adj, int currLen){
        if(currLen > maxLen){
            maxLen = currLen;
            ans = u;
        } else if(currLen == maxLen){
            if(ans > u){
                ans = u;
            }
        }
        for(auto nbr: adj[u]){
            dfs(nbr, adj, currLen+1);
        }
    }


    bool check(string &a, string &big){
        if(a.size() != big.size()-1)    return false;
        for(int i=0; i<a.size(); i++){
            if(a[i] != big[i]){
                return false;
            }
        }
        return true;
    }
};
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