<p>Given a <code>wordlist</code>, we want to implement a spellchecker that converts a query word into a correct word.</p>

<p>For a given <code>query</code> word, the spell checker handles two categories of spelling mistakes:</p>

<ul>
	<li>Capitalization: If the query matches a word in the wordlist (<strong>case-insensitive</strong>), then the query word is returned with the same case as the case in the wordlist.

	<ul>
		<li>Example: <code>wordlist = [&quot;yellow&quot;]</code>, <code>query = &quot;YellOw&quot;</code>: <code>correct = &quot;yellow&quot;</code></li>
		<li>Example: <code>wordlist = [&quot;Yellow&quot;]</code>, <code>query = &quot;yellow&quot;</code>: <code>correct = &quot;Yellow&quot;</code></li>
		<li>Example: <code>wordlist = [&quot;yellow&quot;]</code>, <code>query = &quot;yellow&quot;</code>: <code>correct = &quot;yellow&quot;</code></li>
	</ul>
	</li>
	<li>Vowel Errors: If after replacing the vowels <code>(&#39;a&#39;, &#39;e&#39;, &#39;i&#39;, &#39;o&#39;, &#39;u&#39;)</code> of the query word with any vowel individually, it matches a word in the wordlist (<strong>case-insensitive</strong>), then the query word is returned with the same case as the match in the wordlist.
	<ul>
		<li>Example: <code>wordlist = [&quot;YellOw&quot;]</code>, <code>query = &quot;yollow&quot;</code>: <code>correct = &quot;YellOw&quot;</code></li>
		<li>Example: <code>wordlist = [&quot;YellOw&quot;]</code>, <code>query = &quot;yeellow&quot;</code>: <code>correct = &quot;&quot;</code> (no match)</li>
		<li>Example: <code>wordlist = [&quot;YellOw&quot;]</code>, <code>query = &quot;yllw&quot;</code>: <code>correct = &quot;&quot;</code> (no match)</li>
	</ul>
	</li>
</ul>

<p>In addition, the spell checker operates under the following precedence rules:</p>

<ul>
	<li>When the query exactly matches a word in the wordlist (<strong>case-sensitive</strong>), you should return the same word back.</li>
	<li>When the query matches a word up to capitlization, you should return the first such match in the wordlist.</li>
	<li>When the query matches a word up to vowel errors, you should return the first such match in the wordlist.</li>
	<li>If the query has no matches in the wordlist, you should return the empty string.</li>
</ul>

<p>Given some <code>queries</code>, return a list of words <code>answer</code>, where <code>answer[i]</code> is the correct word for <code>query = queries[i]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> wordlist = ["KiTe","kite","hare","Hare"], queries = ["kite","Kite","KiTe","Hare","HARE","Hear","hear","keti","keet","keto"]
<strong>Output:</strong> ["kite","KiTe","KiTe","Hare","hare","","","KiTe","","KiTe"]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> wordlist = ["yellow"], queries = ["YellOw"]
<strong>Output:</strong> ["yellow"]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= wordlist.length, queries.length &lt;= 5000</code></li>
	<li><code>1 &lt;= wordlist[i].length, queries[i].length &lt;= 7</code></li>
	<li><code>wordlist[i]</code> and <code>queries[i]</code> consist only of only English letters.</li>
</ul>


class Solution {
public:
    vector<string> spellchecker(vector<string>& wordlist, vector<string>& queries) {
		int nq = queries.size(), n = wordlist.size();
		unordered_set<string> dict;
		unordered_map<string, string> lower;
		unordered_map<string, string> vowelMatch;
		set<char> vow;
		vow.insert('a');
		vow.insert('e');
		vow.insert('i');
		vow.insert('o');
		vow.insert('u');
		auto get = [&](const string &w) -> pair<string, string> {
			string r = "", v = "";
			for (char c: w) {
				c = tolower(c);
				r += c;
				if (vow.count(c) == 1) {
					v += '*';
				} else { // chutiya
					v += c;
				}
			}	
			return {r, v};
		};
		for (const string &w: wordlist) {
			dict.insert(w);
			pair<string, string> p = get(w);
			string r = p.first, v = p.second;
			if (lower.count(r) == 0) {
				lower[r] = w;
			}
			if (vowelMatch.count(v) == 0) {
				vowelMatch[v] = w;
			}
		}        
		vector<string> ans(nq);
		for (int i = 0; i < nq; i++) {
			string &s = queries[i];
			if (dict.count(s) == 1) {
				ans[i] = s;
				continue;
			} 
			pair<string, string> p = get(s);
			auto it = lower.find(p.first);
			if (it != lower.end()) {
				ans[i] = it->second;
				continue;
			}
			it = vowelMatch.find(p.second);
			if (it != vowelMatch.end()) {
				ans[i] = it->second;
				continue;
			}
			ans[i] = "";
		}
		return ans;
    }
};

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