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            "title": "Counting Bits",
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            "content": "<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>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 2\n<strong>Output:</strong> [0,1,1]\n<strong>Explanation:</strong>\n0 --&gt; 0\n1 --&gt; 1\n2 --&gt; 10\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 5\n<strong>Output:</strong> [0,1,1,2,1,2]\n<strong>Explanation:</strong>\n0 --&gt; 0\n1 --&gt; 1\n2 --&gt; 10\n3 --&gt; 11\n4 --&gt; 100\n5 --&gt; 101\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= n &lt;= 10<sup>5</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Follow up:</strong></p>\n\n<ul>\n\t<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>\n\t<li>Can you do it without using any built-in function (i.e., like <code>__builtin_popcount</code> in C++)?</li>\n</ul>\n",
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                    "code": "class Solution(object):\n    def countBits(self, n):\n        \"\"\"\n        :type n: int\n        :rtype: List[int]\n        \"\"\"\n        ",
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            "hints": [
                "You should make use of what you have produced already.",
                "Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.",
                "Or does the odd/even status of the number help you in calculating the number of 1s?"
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