#+PROPERTY: STUDY_DECK_02 * TODO 2220. Minimum Bit Flips to Convert Number :easy: :PROPERTIES: :NEETCODE: [[file:../../roadmap.org::*2220. Minimum Bit Flips to Convert Number][2220. Minimum Bit Flips to Convert Number]] :END: A *bit flip* of a number ~x~ is choosing a bit in the binary representation of ~x~ and *flipping* it from either ~0~ to ~1~ or ~1~ to ~0~. - For example, for ~x = 7~, the binary representation is ~111~ and we may choose any bit (including any leading zeros not shown) and flip it. We can flip the first bit from the right to get ~110~, flip the second bit from the right to get ~101~, flip the fifth bit from the right (a leading zero) to get ~10111~, etc. Given two integers ~start~ and ~goal~, return/ the *minimum* number of *bit flips* to convert /~start~/ to /~goal~. *Example 1:* #+begin_src Input: start = 10, goal = 7 Output: 3 Explanation: The binary representation of 10 and 7 are 1010 and 0111 respectively. We can convert 10 to 7 in 3 steps: - Flip the first bit from the right: 1010 -> 1011. - Flip the third bit from the right: 1011 -> 1111. - Flip the fourth bit from the right: 1111 -> 0111. It can be shown we cannot convert 10 to 7 in less than 3 steps. Hence, we return 3. #+end_src *Example 2:* #+begin_src Input: start = 3, goal = 4 Output: 3 Explanation: The binary representation of 3 and 4 are 011 and 100 respectively. We can convert 3 to 4 in 3 steps: - Flip the first bit from the right: 011 -> 010. - Flip the second bit from the right: 010 -> 000. - Flip the third bit from the right: 000 -> 100. It can be shown we cannot convert 3 to 4 in less than 3 steps. Hence, we return 3. #+end_src *Constraints:* - ~0 <= start, goal <= 10^{9}~ *Note:* This question is the same as 461: Hamming Distance. ** TODO Approach Write your approach here. ** TODO Python #+begin_src python class Solution: def minBitFlips(self, start: int, goal: int) -> int: #+end_src ** TODO C++ #+begin_src cpp class Solution { public: int minBitFlips(int start, int goal) { } }; #+end_src