### Problem

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1’s in their binary representation and return them as an array.

#### Example:

For num = 5 you should return [0,1,1,2,1,2].

It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass? Space complexity should be O(n). Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

### Remark

#### Solution

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 `````` ``````class Solution: def countBits(self, num): """ :type num: int :rtype: List[int] """ if num == 0: return  if num == 1: return [0, 1] count = [0, 1] index = 0 flag = 2 for i in range(2, num+1): if i == flag: flag *= 2 index = 0 count.append(count[index]+1) index += 1 return count ``````

 ``````1 2 3 4 5 6 `````` ``````class Solution2: def countBits(self, num): ret =  for i in range(1, num + 1): ret.append(ret[i&(i-1)]+1) return ret ``````