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]
.
Companies:
Google
Related Topics:
String
// OJ: https://leetcode.com/problems/rotated-digits/
// Author: github.com/lzl124631x
// Time: O(ND) where D is the count of digits in N
// Space: O(1)
class Solution {
private:
bool isGood(int N) {
bool good = false;
while (N) {
int d = N % 10;
if (d == 3 || d == 4 || d == 7) return false;
if (d == 2 || d == 5 || d == 6 || d == 9) good = true;
N /= 10;
}
return good;
}
public:
int rotatedDigits(int N) {
int cnt = 0;
for (int i = 1; i <= N; ++i) {
if (isGood(i)) ++cnt;
}
return cnt;
}
};