cpp-flashcards/org/study_deck_02/dsa/bit-manipulation/0338-counting-bits.org

#+ANKI_DECK: study_deck_02
* TODO 0338. Counting Bits :easy:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*0338. Counting Bits][0338. Counting Bits]]
:END:

Given an integer ~n~, return /an array /~ans~/ of length /~n + 1~/ such that for each /~i~/ /(~0 <= i <= n~)/, /~ans[i]~/ is the *number of */~1~/*'s* in the binary representation of /~i~.

*Example 1:*


#+begin_src
Input: n = 2
Output: [0,1,1]
Explanation:
0 --> 0
1 --> 1
2 --> 10
#+end_src


*Example 2:*


#+begin_src
Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101
#+end_src


*Constraints:*

 - ~0 <= n <= 10^{5}~

*Follow up:*

 - It is very easy to come up with a solution with a runtime of ~O(n log n)~. Can you do it in linear time ~O(n)~ and possibly in a single pass?

 - Can you do it without using any built-in function (i.e., like ~__builtin_popcount~ in C++)?

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python :lc-problem 338 :lc-lang python3
class Solution:
    def countBits(self, n: int) -> List[int]:
#+end_src

** TODO C++
#+begin_src cpp :lc-problem 338
class Solution {
public:
    vector<int> countBits(int n) {
        
    }
};
#+end_src