## Question

An integer `x`

is a **good** 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. For example:

`0`

,`1`

, and`8`

rotate to themselves,`2`

and`5`

rotate to each other (in 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.

Given an integer `n`

, return *the number of good integers in the range *

`[1, n]`

.**Example 1:**

Input:n = 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.

**Example 2:**

Input:n = 1Output:0

**Example 3:**

Input:n = 2Output:1

**Constraints:**

`1 <= n <= 10`

^{4}

## Python Solution

class Solution: def rotatedDigits(self, N: int) -> int: d = {'0':'0' , '1':'1' , '2':'5' , '8':'8' , '5':'2' , '6':'9' , '9':'6' } def checker(n): k='' for i in str(n): if i not in d.keys(): return False else: k+=d[i] if k!=str(n): return True else: return False l=0 for i in range(N+1): if checker(i): l+=1 return l