[LeetCode Sync] Runtime - 3 ms (94.01%), Memory - 18.5 MB (59.13%)

Author: NinePiece2

Solutions/0338-counting-bits/README.md

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+<p>Given an integer <code>n</code>, return <em>an array </em><code>ans</code><em> of length </em><code>n + 1</code><em> such that for each </em><code>i</code><em> </em>(<code>0 &lt;= i &lt;= n</code>)<em>, </em><code>ans[i]</code><em> is the <strong>number of </strong></em><code>1</code><em><strong>&#39;s</strong> in the binary representation of </em><code>i</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 2
+<strong>Output:</strong> [0,1,1]
+<strong>Explanation:</strong>
+0 --&gt; 0
+1 --&gt; 1
+2 --&gt; 10
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 5
+<strong>Output:</strong> [0,1,1,2,1,2]
+<strong>Explanation:</strong>
+0 --&gt; 0
+1 --&gt; 1
+2 --&gt; 10
+3 --&gt; 11
+4 --&gt; 100
+5 --&gt; 101
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>0 &lt;= n &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Follow up:</strong></p>
+
+<ul>
+	<li>It is very easy to come up with a solution with a runtime of <code>O(n log n)</code>. Can you do it in linear time <code>O(n)</code> and possibly in a single pass?</li>
+	<li>Can you do it without using any built-in function (i.e., like <code>__builtin_popcount</code> in C++)?</li>
+</ul>

Solutions/0338-counting-bits/solution.py

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+class Solution:
+    def countBits(self, n: int) -> List[int]:
+        return [i.bit_count() for i in range(n + 1)]