<filename>easy/788-Rotated Digits.py
"""
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: 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.
Note:

N  will be in range [1, 10000].
"""

# time complexity: O(n), space complexity: O(n)

class Solution:
    def rotatedDigits(self, N: int) -> int:
        return sum(list(self.check(i) for i in range(1, N + 1)))

    def check(self, number: int) -> bool:
        dic = {0: 0, 1: 1, 8: 8, 2: 5, 5: 2, 6: 9, 9: 6}
        newnumber = ''
        oldnumber = number
        while number > 0:
            remainder = number % 10
            if remainder not in dic:
                return 0
            newnumber = str(dic[remainder]) + newnumber
            number //= 10
        return int(int(newnumber) != oldnumber)


<filename>easy/788-Rotated Digits.py
"""
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: 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.
Note:

N  will be in range [1, 10000].
"""

# time complexity: O(n), space complexity: O(n)

class Solution:
    def rotatedDigits(self, N: int) -> int:
        return sum(list(self.check(i) for i in range(1, N + 1)))

    def check(self, number: int) -> bool:
        dic = {0: 0, 1: 1, 8: 8, 2: 5, 5: 2, 6: 9, 9: 6}
        newnumber = ''
        oldnumber = number
        while number > 0:
            remainder = number % 10
            if remainder not in dic:
                return 0
            newnumber = str(dic[remainder]) + newnumber
            number //= 10
        return int(int(newnumber) != oldnumber)


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: 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. Note:  N  will be in range [1, 10000]. # time complexity: O(n), space complexity: O(n)