Merge pull request #19 from iamAntimPal/Leetcode-75

Leetcode 75

Author: iamAntimPal

Solution/338. Counting Bits/338. Counting Bits.py

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

Solution/338. Counting Bits/readme.md

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+---
+comments: true
+difficulty: Easy
+edit_url: 
+tags:
+    - Bit Manipulation
+    - Dynamic Programming
+---
+
+<!-- problem:start -->
+
+# [338. Counting Bits](https://leetcode.com/problems/counting-bits)
+
+
+## Description
+
+<!-- description:start -->
+
+<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>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def countBits(self, n: int) -> List[int]:
+        return [i.bit_count() for i in range(n + 1)]
+```
+
+#### Java
+
+```java
+class Solution {
+    public int[] countBits(int n) {
+        int[] ans = new int[n + 1];
+        for (int i = 0; i <= n; ++i) {
+            ans[i] = Integer.bitCount(i);
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    vector<int> countBits(int n) {
+        vector<int> ans(n + 1);
+        for (int i = 0; i <= n; ++i) {
+            ans[i] = __builtin_popcount(i);
+        }
+        return ans;
+    }
+};
+```
+
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->