<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>


bool compareBySize(const std::string &a, const std::string &b) {
    if (a.size() == b.size()){
        return a < b;
    }
    return a.size() < b.size();
}

class Solution {
public:
    string longestWord(vector<string>& words) {
        sort(words.begin(), words.end(), compareBySize);
        unordered_set<string> a;
        int n=words.size(), i=0, count=0;
        string temp, ans="";
        for(i=0;i<n;i++){
            if(words[i].size()==1){
                a.insert(words[i]);
                count=1;
            }
            else{break;}
        }
        if(count!=1){return ans;}
        ans=words[0];
        
        while(i<n){
            if(words[i].size() == count+1){
                temp = words[i];
                temp.pop_back();
                if(a.find(temp)!=a.end()){
                    ans=words[i];
                    count++;
                    a.insert(words[i]);
                }
            }
            else if(words[i].size() == count){
                temp = words[i];
                temp.pop_back();
                if(a.find(temp)!=a.end()){
                    a.insert(words[i]);
                }
            }
            else{break;}
            i++;
        }
        return ans;
    }
};
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