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?
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].
1 class Solution {
2 public int rotatedDigits(int N) {
3 int count = 0;
4 for (int i = 1; i <= N; i ++) {
5 if (isValid(i)) count ++;
6 }
7 return count;
8 }
9
10 public boolean isValid(int N) {
11 /*
12 Valid if N contains ATLEAST ONE 2, 5, 6, 9
13 AND NO 3, 4 or 7s
14 */
15 boolean validFound = false;
16 while (N > 0) {
17 if (N % 10 == 2) validFound = true;
18 if (N % 10 == 5) validFound = true;
19 if (N % 10 == 6) validFound = true;
20 if (N % 10 == 9) validFound = true;
21 if (N % 10 == 3) return false;
22 if (N % 10 == 4) return false;
23 if (N % 10 == 7) return false;
24 N = N / 10;
25 }
26 return validFound;
27 }
28 }
