> [!META]- Inline Metadata
> [tags:: #career #concepts/programming/bit-manipulation ]
> [project group:: Neetcode 150]
> [difficulty:: easy]
> [status:: done]
> [link:: https://leetcode.com/problems/reverse-bits/submissions/]
> [up:: [[Leetcode MOC]]]
## Problem
Reverse bits of a given 32 bits unsigned integer.
**Note:**
- Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
- In Java, the compiler represents the signed integers using [2's complement notation](https://en.wikipedia.org/wiki/Two%27s_complement). Therefore, in **Example 2** above, the input represents the signed integer `-3` and the output represents the signed integer `-1073741825`.
**Example 1:**
**Input:** n = 00000010100101000001111010011100
**Output:** 964176192 (00111001011110000010100101000000)
**Explanation:** The input binary string **00000010100101000001111010011100** represents the unsigned integer 43261596, so return 964176192 which its binary representation is **00111001011110000010100101000000**.
**Example 2:**
**Input:** n = 11111111111111111111111111111101
**Output:** 3221225471 (10111111111111111111111111111111)
**Explanation:** The input binary string **11111111111111111111111111111101** represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is **10111111111111111111111111111111**.
**Constraints:**
- The input must be a **binary string** of length `32`
**Follow up:** If this function is called many times, how would you optimize it?
## Solution
### Mine
#### Python
```python
class Solution:
def reverseBits(self, n: int) -> int:
return int(f"{n:032b}"[::-1], 2)
```
#### Rust
### Neetcode
```python
class Solution:
def reverseBits(self, n: int) -> int:
res = 0
for i in range(32):
bit = (n >> i) & 1
res = res | (bit << (31 - i))
return res
```
## Things I Learned
### Neetcode
For each bit, take `(n >> i)` to get the bit we want in the ones spot. Then shift it to the left by 31 - 1 and log or it with `res`.
## Scratchpad
The obvious solution would be to just get the string representation, flip it, and convert back to an int. I suspect there's a technique that's more performant and whatnot.