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].

• 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
```

Share It, If You Like It.