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# Reverse Number

Given a signed 32-bit integer `x`, return `x` with its digits reversed. If reversing `x` causes the value to go outside the signed 32-bit integer range `[-2^31, 2^31 - 1]`, then return `0`.

**Assume the environment does not allow you to store 64-bit integers (signed or unsigned).**

```
Example 1:

Input: x = 123
Output: 321
Example 2:

Input: x = -123
Output: -321
Example 3:

Input: x = 120
Output: 21

```

#### Constraints:

`-231 <= x <= 231 - 1`

### Solution

```csharp
public int Reverse(int x) {
    long result = 0;
    while (x != 0) {
        result = result * 10 + x % 10;
        x /= 10;
        if (result > Int32.MaxValue || result < Int32.MinValue) {
            return 0;
        }
    }
    return (int)result;
}

```

The time complexity of the algorithm is **O(log(x))**, where x is the input to the function. This is because in each iteration of the while loop, we’re dividing `x` by `10`, effectively reducing the problem size by a factor of `10`. Hence, the number of iterations is proportional to the number of digits in x, which is log(x) in base 10.

The space complexity of the algorithm is **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 x are the only variables being used, and their size does not change with different inputs.


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