X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X. 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; 6 and 9 rotate to each other, and the rest of the numbers do not rotate to any other number.
Now given a positive number N, how many numbers X from 1 to N are good?
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.
N will be in range [1, 10000].
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
## Blog link: https://code.dennyzhang.com/rotated-digits ## Basic Ideas: Check number one by one ## ## Complexity: Time O(n), Space O(1) class Solution: def rotatedDigits(self, N): """ :type N: int :rtype: int """ res = 0 for num in range(1, N+1): has_changed, is_valid = False, True for ch in str(num): if ch in "347": is_valid = False break if ch in "2569": has_changed = True if has_changed and is_valid: res += 1 return res
Original URL: https://code.dennyzhang.com/rotated-digits