### Number of Even and Odd Bits

You are given a **positive** integer `n`.

Let `even` denote the number of even indices in the binary representation of `n` (**0-indexed**) with value `1`.

Let `odd` denote the number of odd indices in the binary representation of `n` (**0-indexed**) with value `1`.

Return *an integer array* `answer` *where* `answer = [even, odd]`.

 

**Example 1:**

```
Input: n = 17
Output: [2,0]
Explanation: The binary representation of 17 is 10001. 
It contains 1 on the 0th and 4th indices. 
There are 2 even and 0 odd indices.
```

**Example 2:**

```
Input: n = 2
Output: [0,1]
Explanation: The binary representation of 2 is 10.
It contains 1 on the 1st index. 
There are 0 even and 1 odd indices.
```

 

**Constraints:**

- `1 <= n <= 1000`


**Function definition**

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