# LeetCode 338. Counting Bits

## Description

https://leetcode.com/problems/counting-bits/

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`

• 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++)?

## Explanation

Convert number to binary strings and count ‘1’s.

## Python Solution

``````class Solution:
def countBits(self, n: int) -> List[int]:

results = []

for i in range(n + 1):
bin_str = bin(i)[2:]

one_count = bin_str.count('1')
results.append(one_count)

return results    ``````
• Time Complexity: O(N).
• Space Complexity: O(N).