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# Reverse Number II

Reverse bits of a given 32 bits unsigned integer.

```
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`.

### Solution

```csharp
public uint ReverseBits(uint n) {
    uint result = 0;
    for (int i = 0; i < 32; i++) {
        result <<= 1;
        if ((n & 1) == 1)
            result++;
        n >>= 1;
    }
    return result;
}

```

This function works by iterating through the bits of `n` from right to left. In each iteration, it shifts `result` to the left to make room for the next bit, and then adds the least significant bit of `n` to result. It then shifts `n` to the right to process the next bit. This continues until all bits of `n` have been processed.

The time complexity of the algorithm is **O(1)**. This is because the number of iterations of the loop is fixed at `32`, regardless of the input.

The space complexity of the algorithm is also **O(1)**, which means it uses a constant amount of space. This is because we’re only using a fixed number of variables and not any data structures that grow with the size of the input. The integer result and the input n are the only variables being used, and their size does not change with different inputs.