<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>

class Solution:
    def countBits(self, n: int) -> List[int]:
        l = [0] * (n + 1)
        for i in range(1, n + 1):
            l[i] = l[i >> 1] + (i & 1)
        return l

23100