Counting bits

Time: O(N); Space: O(N); medium

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: num = 2

Output: [0,1,1]

Example 2:

Input: num = 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.

Hints:

  1. You should make use of what you have produced already.

  2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.

  3. Or does the odd/even status of the number help you in calculating the number of 1s?

[3]:

class Solution1(object):
    def countBits(self, num) -> list:
        """
        :type num: int
        :rtype: List[int]
        """
        res = [0]
        for i in range(1, num + 1):
            # Number of 1's in i = (i & 1) + number of 1's in (i / 2).
            res.append((i & 1) + res[i >> 1])
        return res

[4]:

s = Solution1()
num = 2
assert s.countBits(num) == [0,1,1]
num = 5
assert s.countBits(num) == [0,1,1,2,1,2]

[5]:

class Solution2(object):
    def countBits(self, num) -> list:
        """
        :type num: int
        :rtype: List[int]
        """
        s = [0]
        while len(s) <= num:
            s.extend(list(map(lambda x: x + 1, s)))
        # print(s)      # [0, 1, 1, 2, 1, 2, 2, 3]
        return s[:num + 1]

[6]:

s = Solution2()
num = 2
assert s.countBits(num) == [0,1,1]
num = 5
assert s.countBits(num) == [0,1,1,2,1,2]