[LeetCode] Challenge log 788

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788. Rotated Digits

It is a logic problem. My solution is:

  1. If it contains 3 4 7, it is invalid.
  2. If it contains only 0 1 8, it is valid but not different.
  3. The rest cases are passed.

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; 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?

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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].
Soulution:
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class Solution:
    def rotatedDigits(self, N):
        """
        :type N: int
        :rtype: int
        """
        res = 0
        for i in range(1, N+1):
            num = set(str(i))
            if len(num & {'3', '4', '7'}) > 0:
                continue
            elif num & {'0', '1', '8'} == num:
                continue
            else:
                res += 1
        return res
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