788. Rotated Digits

1. Description

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?

2. Example

Example 1:
Input: 10
Output: 4
Explanation:
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.

3. Note

• N will be in range [1, 10000].

4. Solutions

My Accepted Solution

n = number
Time complexity: O($nlog_{10}n$)
Space complexity: O(1)

class Solution 
{
    bool isInvalidDigit(char digit)
    {
        return (digit == '3' || digit == '4' || digit == '7');
    }
    
    bool isDifferentDigit(char digit)
    {
        return (digit == '2' || digit == '5' || digit == '6' || digit == '9');
    }
    
    bool isNumberValid(int number)
    {
        string digits(to_string(number));
        int differentDigitsCount = 0;
        for(int i = 0; i < digits.size(); i++)
        {
            if(isInvalidDigit(digits[i])) return false;
            
            if(isDifferentDigit(digits[i])) differentDigitsCount++;
        }
        
        return (differentDigitsCount > 0);
    }
public:
    // int rotatedDigits(int N) 
    int rotatedDigits(int number) 
    {
        int result = 0;
        for(int i = 1; i <= number; i++)
        {
            if(isNumberValid(i))
                result++;
        }
        
        return result;
    }
};