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Author: NinePiece2

Solutions/0837-most-common-word/README.md

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+<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>

Solutions/0837-most-common-word/solution.py

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+class Solution:
+    def mostCommonWord(self, paragraph: str, banned: List[str]) -> str:
+        banned_Set = set(banned)
+        count = Counter(re.findall("[a-z]+", paragraph.lower()))
+        return next(word for word, _ in count.most_common() if word not in banned_Set)