## Description

https://leetcode.com/problems/rotated-digits/

`x`

is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from `x`

. Each digit must be rotated – we cannot choose to leave it alone.

A number is valid if each digit remains a digit after rotation. 0, 1, and 8 rotate to themselves; 2 and 5 rotate to each other (on this case they are rotated in a different direction, in other words 2 or 5 gets mirrored); 6 and 9 rotate to each other, and the rest of the numbers do not rotate to any other number and become invalid.

Now given a positive number `n`

, how many numbers `x`

from `1`

to `n`

are good?

Example:Input:10Output:4Explanation:There are four good numbers in the range [1, 10] : 2, 5, 6, 9. Note that 1 and 10 are not good numbers, since they remain unchanged after rotating.

**Note:**

`n`

will be in range`[1, 10000]`

.

## Explanation

From 1 to N, check how many numbers can have different rotate numbers.

## Python Solution

```
class Solution:
def rotatedDigits(self, N: int) -> int:
count = 0
mapping = {
'0': '0',
'1': '1',
'2': '5',
'5': '2',
'6': '9',
'8': '8',
'9': '6',
}
for i in range(1, N + 1):
i_str = str(i)
rotated_str = ""
can_rotate = True
for c in i_str:
if c not in mapping:
can_rotate = False
break
rotated_str += mapping[c]
if can_rotate and rotated_str != i_str:
count += 1
return count
```

- Time Complexity: O(N).
- Space Complexity: O(N).