Day-28_Counting_Bits.py

'''
	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.
   Hide Hint #1  
You should make use of what you have produced already.
   Hide Hint #2  
Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
   Hide Hint #3  
Or does the odd/even status of the number help you in calculating the number of 1s?


'''

class Solution:
    def countBits(self, num):
        dp = [0] * (num + 1)
        for pow in range(0, 32):
            start, end = 1<<pow, 1<<(pow + 1)
            if start > num: break

            for j in range(start, min(num+1,end)):
                dp[j] = dp[j-start] + 1
        return dp