338 Counting Bits

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: Fornum = 5you should return[0,1,1,2,1,2].

Follow up:

• 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.

The Idea: This is an identified sequence, (see https://oeis.org/A000120). The pattern can be recursively identified. One way it is identified is through this triangular pattern.

  0,
  1, <- base case
  1,2,
  1,2,2,3,
  1,2,2,3,2,3,3,4,
  1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
  1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,

Beginning with the two base cases 0 and 1, we can identify the pattern simply copies the layer above it twice with the second half incremented by 1. The pattern grows by 2^n, but we can always return a subvector [0:num] since 2^n >= num.

Complexity: O(num) time and O(log(num)) extra space.

def __countBits(self, A000120, prev, num):
    """
    :param num: int
    :return: List[int]
    """

    if len(A000120) >= num + 1:
        return A000120
    else:
        A000120.extend(prev)
        double = [i+1 for i in prev]
        A000120.extend(double)
        prev.extend(double)
        return self.__countBits(A000120, prev, num)

def countBits(self, num):
    """
    :type num: int
    :rtype: List[int]
    """

    if num == 0: return [0]
    elif num == 1: return [0,1]

    return self.__countBits([0,1], [1], num)[0:num+1]