<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>

// Define the Trie Node structure
struct TrieNode {
    unordered_map<char, TrieNode*> children;
    bool endOfWord = false;
};

class Solution {
public:
    // Insert a word into the Trie
    void insert(TrieNode* root, const string& word) {
        TrieNode* node = root;
        for (char ch : word) {
            if (!node->children.count(ch)) {
                node->children[ch] = new TrieNode();
            }
            node = node->children[ch];
        }
        node->endOfWord = true;  // Mark the end of the word
    }

    // DFS to find the longest word in the Trie
    void dfs(TrieNode* node, string& currentWord, string& result) {
        for (auto& [ch, childNode] : node->children) {
            if (childNode->endOfWord) {
                currentWord.push_back(ch);

                // Update the result if currentWord is longer or lexicographically smaller
                if (currentWord.length() > result.length() ||
                    (currentWord.length() == result.length() && currentWord < result)) {
                    result = currentWord;
                }

                // Continue DFS on the child node
                dfs(childNode, currentWord, result);

                // Backtrack
                currentWord.pop_back();
            }
        }
    }

    string longestWord(vector<string>& words) {
        TrieNode* root = new TrieNode();

        // Build the Trie
        for (const string& word : words) {
            insert(root, word);
        }

        string result = "";
        string currentWord = "";
        dfs(root, currentWord, result);  // Perform DFS to find the longest word

        return result;
    }
};

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