338_counting_bits.py

'''
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 1:

Input: 2
Output: [0,1,1]
Example 2:

Input: 5
Output: [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.

def countBits(self, num: int) -> List[int]:
'''


class Solution:
    def countBits(self, num):
        '''
        approach:
            dp + least significant bit
        '''
        dp = [0]
        for i in range(1, num+1):
            # dp.append(dp[i//2]+(i % 2))
            dp.append(dp[i >> 1] + (i & 1))
        return dp


if __name__ == '__main__':
    # begin
    s = Solution()
    print(s.countBits(5))