LeetCode Link: https://leetcode.com/problems/rotated-digits/
Language: C#
Problem Statement
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].
Examples
Example 1:
Input: n = 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.
Example 2:
Input: n = 1
Output: 0
Example 3:
Input: n = 2
Output: 1
Constraints
- 1 <= n <= 104
Solution
public class Solution
{
public int RotatedDigits(int n)
{
int count = 0;
while (n > 0)
{
count += IsGood(n--) ? 1 : 0;
}
return count;
}
private bool IsGood(int x)
{
int i;
int y = x;
bool rotated = false;
while (x>0)
{
i = x%10;
switch(i)
{
case 0:
case 1:
case 8:
break;
case 2:
case 5:
case 6:
case 9:
rotated = true;
break;
default:
return false;
}
x /= 10;
}
return rotated;
}
}
Complexity
Time Complexity: O(N*L)
where N is the range and L is the number of digits.
Space Complexity: O(1)
Notes
This can also be solved using Dynamic Programming. I have not used it in this solution.