Leetcode: Counting Bits

Counting Bits



Similar Problems:


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:
For num = 5 you 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.

Github: code.dennyzhang.com

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 [0]
        if num == 1:
            return [0, 1]
        res = [0]*(num+1)
        res[0] = 0
        res[1] = 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
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