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43 lines
1.3 KiB
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43 lines
1.3 KiB
HTML
<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 <= i <= n</code>)<em>, </em><code>ans[i]</code><em> is the <strong>number of </strong></em><code>1</code><em><strong>'s</strong> in the binary representation of </em><code>i</code>.</p>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> n = 2
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<strong>Output:</strong> [0,1,1]
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<strong>Explanation:</strong>
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0 --> 0
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1 --> 1
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2 --> 10
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> n = 5
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<strong>Output:</strong> [0,1,1,2,1,2]
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<strong>Explanation:</strong>
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0 --> 0
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1 --> 1
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2 --> 10
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3 --> 11
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4 --> 100
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5 --> 101
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>0 <= n <= 10<sup>5</sup></code></li>
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</ul>
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<p> </p>
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<p><strong>Follow up:</strong></p>
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<ul>
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<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>
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<li>Can you do it without using any built-in function (i.e., like <code>__builtin_popcount</code> in C++)?</li>
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</ul>
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