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Kotlin-based LeetCode algorithm problem solutions, regularly updated
338\. Counting Bits
Easy
Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** [0,1,1]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n = 5
**Output:** [0,1,1,2,1,2]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101
**Constraints:**
* 0 <= n <= 105
**Follow up:**
* It is very easy to come up with a solution with a runtime of `O(n log n)`. Can you do it in linear time `O(n)` and possibly in a single pass?
* Can you do it without using any built-in function (i.e., like `__builtin_popcount` in C++)?