<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 {
    private:
    static bool cmp(string& s1, string& s2){
        if(s1.length() == s2.length()){
            return s1 > s2;
        }
        return s1.length() < s2.length();
    }

    public:
    string longestWord(vector<string>& words) {
        unordered_set<string> uset; // Unordered set
        for(int i = 0; i < words.size(); i++){
            uset.insert(words[i]);
        }
        vector<string> arr;
        for(int i = 0; i < words.size(); i++){
            int check = 0;
            for(int j = 0; j <= words[i].length() - 1; j++){
                if(uset.find(words[i].substr(0, j + 1)) == uset.end()){
                    check = 1;
                    break;
                }
            }
            if(check == 0){
                arr.push_back(words[i]);
            }
        }
        if(arr.empty()){
            return "";
        }
        sort(arr.begin(), arr.end(), cmp);
        return arr[arr.size() - 1];
    }
};
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