Most Common Word
<p>Given a string <code>paragraph</code> and a string array of the banned words <code>banned</code>, return <em>the most frequent word that is not banned</em>. It is <strong>guaranteed</strong> there is <strong>at least one word</strong> that is not banned, and that the answer is <strong>unique</strong>.</p>

<p>The words in <code>paragraph</code> are <strong>case-insensitive</strong> and the answer should be returned in <strong>lowercase</strong>.</p>

<p><strong>Note</strong> that words can not contain punctuation symbols.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> paragraph = &quot;Bob hit a ball, the hit BALL flew far after it was hit.&quot;, banned = [&quot;hit&quot;]
<strong>Output:</strong> &quot;ball&quot;
<strong>Explanation:</strong> 
&quot;hit&quot; occurs 3 times, but it is a banned word.
&quot;ball&quot; occurs twice (and no other word does), so it is the most frequent non-banned word in the paragraph. 
Note that words in the paragraph are not case sensitive,
that punctuation is ignored (even if adjacent to words, such as &quot;ball,&quot;), 
and that &quot;hit&quot; isn&#39;t the answer even though it occurs more because it is banned.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> paragraph = &quot;a.&quot;, banned = []
<strong>Output:</strong> &quot;a&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= paragraph.length &lt;= 1000</code></li>
	<li>paragraph consists of English letters, space <code>&#39; &#39;</code>, or one of the symbols: <code>&quot;!?&#39;,;.&quot;</code>.</li>
	<li><code>0 &lt;= banned.length &lt;= 100</code></li>
	<li><code>1 &lt;= banned[i].length &lt;= 10</code></li>
	<li><code>banned[i]</code> consists of only lowercase English letters.</li>
</ul>
Array; Hash Table; String; Counting
function mostCommonWord(paragraph: string, banned: string[]): string {
    const s = paragraph.toLocaleLowerCase();
    const map = new Map<string, number>();
    const set = new Set<string>(banned);
    for (const word of s.split(/[^A-z]/)) {
        if (word === '' || set.has(word)) {
            continue;
        }
        map.set(word, (map.get(word) ?? 0) + 1);
    }
    return [...map.entries()].reduce((r, v) => (v[1] > r[1] ? v : r), ['', 0])[0];
}
