<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:
    unordered_set<string> word_set;
    unordered_map<string, bool> dp;

    bool dfs(string &word)
    {
        if(dp.find(word) != dp.end())
            return dp[word];

        if (word.size() == 1)
        {
            dp[word] = true;
            return true;
        }
        else
        {
            string prefix = word.substr(0, word.length() - 1);
            if (word_set.find(prefix) != word_set.end())
            {
                dp[word] = dfs(prefix);
                return dp[word];
            }
            else
            {
                dp[word] = false;
                return false;
            }
        }
    }

    string longestWord(vector<string> &words)
    {
        word_set.insert(words.begin(), words.end());

        int max_len = 0;
        string ans_word = "";
        for (string word : words)
        {
            if (!dfs(word))
                continue;

            if (word.length() > max_len)
            {
                max_len = word.length();
                ans_word = word;
            }
            else if (word.length() == max_len)
            {
                if (ans_word > word)
                    ans_word = word;
            }
        }

        return ans_word;
    }
};
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