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.
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.
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
## Blog link: https://code.dennyzhang.com/counting-bits ## Basic Ideas: ## If k%2 == 0, f(k) = f(k/2) ## If k%2 == 1, f(k) = f(k-1) ## ## Complexity: Time O(n), Space O(n) class Solution(object): def countBits(self, num): """ :type num: int :rtype: List[int] """ if num == 0: return  if num == 1: return [0, 1] res = *(num+1) res = 0 res = 1 for i in range(2, num+1): if i%2 == 0: res[i] = res[i/2] else: res[i] = res[i-1]+1 return res