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# LC190 - Reverse Bits

## 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

**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**.

## Solution

### Intuition

Iterate in reverse and multiply each character by 2 to the power of i. The main point of complexity is input handling, which is formatted as an integer in python. Hence, we simply convert it to a string, keeping the leading zeroes with `format(n, '032b`

```python
def reverseBits(self, n: int) -> int:
	res = 0
	nstr = format(n, '032b')
	for i in range(31, -1, -1):
		res += int(nstr[i], pow(2, i))
	return res
```

### Optimizing

Use fully bitwise operations and eliminate string conversion.

```python
def reverseBits(self, n: int) -> int:
	res = 0
	for i in range(32):
		bit = (n >> i) & 1
		res = res | (bit << (31-i))
	return res
```