<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 TrieNode {
public:
    bool end; // if node is last char of a word
    vector<TrieNode*> leaves; // array of all letters
    TrieNode() {
        leaves = vector<TrieNode*>(26, nullptr); // declare alphabet
        end = false; // default as is not end of word
    }
};
class Solution {
private:
    TrieNode* root; // root array with alphabet
public:
    string longestWord(vector<string>& words) {
        root = new TrieNode(); // initialize root
        root -> end = true; // mark root as end of word for linear (see search)
        string output = ""; // declare output

        for (string& w : words) {
            insert(w); // insert all words into trie
        }

        for (string& w : words) { // read all words again and check if all substrs are in dict + is greater than output
            if (compareWords(w, output) && search(w)) { // compare first since is O 1 and search is O n
                output = w;
            }
        }
        return output;
    }

    void insert(string& word) { // insert word into trie
        TrieNode* curr = root; // dummy
        for (char& c : word) { // read word and add each char
            if (!curr -> leaves[c - 'a']) curr -> leaves[c - 'a'] = new TrieNode(); // create new node if null
            curr = curr -> leaves[c - 'a']; // traverse word
        }
        curr -> end = true; // mark last char as end of word
    }

    bool search(string word) {
        TrieNode* curr = root; // dummy
        for (char& c : word) { // read word
            if (!curr -> end) return false; // for this, all nodes need to be end of a word
            if (!curr -> leaves[c - 'a']) return false; // and they need to exist aswell (leave, next char)
            curr = curr -> leaves[c - 'a']; // traverse
        }
        return curr -> end; // return if last char is end of word (it should be since we search for words weve added)
    }

    bool compareWords(const string& w1, const string& w2) { // compared based on questions sprecifications / reqs
        return w1.size() > w2.size() || (w1.size() == w2.size() && w1 < w2);
    }
};
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