## Description

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]


• 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.

## Solutions

### 1. Brute Force

# 1. Brute Force
# Time: O(num)
# Space: O(num)
class Solution:
def countBits(self, num: int) -> List[int]:
res = []
for i in range(num+1):
cnt = 0
while i != 0:
div, mod = divmod(i, 2)
if mod == 1:
cnt += 1
i = div
res.append(cnt)
return res

# 15/15 cases passed (300 ms)
# Your runtime beats 5.01 % of python3 submissions
# Your memory usage beats 5 % of python3 submissions (19.5 MB)


### 2. DP

状态转移矩阵：f[i] = f[i / 2] + i % 2.

# Time: O(num)
# Space: O(num)
class Solution:
def countBits(self, num: int) -> List[int]:
res =  * (num + 1)
for i in range(num + 1):
res[i] = res[i >> 1] + i % 2
return res
# 15/15 cases passed (88 ms)
# Your runtime beats 56.5 % of python3 submissions
# Your memory usage beats 5 % of python3 submissions (19.6 MB)