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Author: NinePiece2

Solutions/0899-binary-gap/README.md

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+<p>Given a positive integer <code>n</code>, find and return <em>the <strong>longest distance</strong> between any two <strong>adjacent</strong> </em><code>1</code><em>&#39;s in the binary representation of </em><code>n</code><em>. If there are no two adjacent </em><code>1</code><em>&#39;s, return </em><code>0</code><em>.</em></p>
+
+<p>Two <code>1</code>&#39;s are <strong>adjacent</strong> if there are only <code>0</code>&#39;s separating them (possibly no <code>0</code>&#39;s). The <b>distance</b> between two <code>1</code>&#39;s is the absolute difference between their bit positions. For example, the two <code>1</code>&#39;s in <code>&quot;1001&quot;</code> have a distance of 3.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 22
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 22 in binary is &quot;10110&quot;.
+The first adjacent pair of 1&#39;s is &quot;<u>1</u>0<u>1</u>10&quot; with a distance of 2.
+The second adjacent pair of 1&#39;s is &quot;10<u>11</u>0&quot; with a distance of 1.
+The answer is the largest of these two distances, which is 2.
+Note that &quot;<u>1</u>01<u>1</u>0&quot; is not a valid pair since there is a 1 separating the two 1&#39;s underlined.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 8
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> 8 in binary is &quot;1000&quot;.
+There are not any adjacent pairs of 1&#39;s in the binary representation of 8, so we return 0.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 5
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 5 in binary is &quot;101&quot;.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
+</ul>

Solutions/0899-binary-gap/solution.py

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+class Solution:
+    def binaryGap(self, n: int) -> int:
+        result, last, current = 0, float('inf'), 0
+        while n:
+            if n & 1:
+                result = max(result, current - last)
+                last = current
+            current += 1
+            n >>= 1
+        return result