---
# System prepended metadata

title: Leetcode 190. Reverse Bits
tags: [Leetcode]

---

[link](https://leetcode.com/problems/reverse-bits/)

---

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. 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

---

The variable res is initialized to 0, which will store the result. The code enters a for loop that iterates from 0 to 31, representing the 32 bits in the integer.

Inside the loop, the code extracts the i-th bit from n by performing a right shift and bitwise AND operation: (n >> i) & 1. This gives the value of the i-th bit (either 0 or 1). The extracted bit is then placed in the reversed position by performing a left shift: bit << (31 - i). This moves the bit to the (31 - i)-th position from the left.

The result res is updated by performing a bitwise OR operation between res and the shifted bit: res = res | (bit << (31 - i)). This sets the i-th bit of res to the extracted bit value. After completing the loop, res contains the reversed bits of n.

#### Solution 1
```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
```
O(T): O(1)
O(S): O(1)
