cpp-flashcards/org/study_deck_02/dsa/bit-manipulation/0231-power-of-two.org

#+ANKI_DECK: study_deck_02
* TODO 0231. Power of Two :easy:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*0231. Power of Two][0231. Power of Two]]
:END:

Given an integer ~n~, return /~true~ if it is a power of two. Otherwise, return ~false~/.

An integer ~n~ is a power of two, if there exists an integer ~x~ such that ~n == 2^{x}~.

*Example 1:*


#+begin_src
Input: n = 1
Output: true
Explanation: 20 = 1
#+end_src


*Example 2:*


#+begin_src
Input: n = 16
Output: true
Explanation: 24 = 16
#+end_src


*Example 3:*


#+begin_src
Input: n = 3
Output: false
#+end_src


*Constraints:*

 - ~-2^{31} <= n <= 2^{31} - 1~

*Follow up:* Could you solve it without loops/recursion?

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python :lc-problem 231 :lc-lang python3
class Solution:
    def isPowerOfTwo(self, n: int) -> bool:
#+end_src

** TODO C++
#+begin_src cpp :lc-problem 231
class Solution {
public:
    bool isPowerOfTwo(int n) {
        
    }
};
#+end_src