338 Counting Bits
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.
num = 5you should return
- 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.
1, <- base case
Beginning with the two base cases 0 and 1, we can identify the pattern simply copies the layer above it twice with the second half incremented by 1. The pattern grows by
2^n, but we can always return a subvector
2^n >= num.
Complexity: O(num) time and O(log(num)) extra space.
def __countBits(self, A000120, prev, num):
:param num: int
if len(A000120) >= num + 1:
double = [i+1 for i in prev]
return self.__countBits(A000120, prev, num)
def countBits(self, num):
:type num: int
if num == 0: return 
elif num == 1: return [0,1]
return self.__countBits([0,1], , num)[0:num+1]