Formatted question description: https://leetcode.ca/all/788.html

788. Rotated Digits (Easy)

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

Solution 1.

// OJ: https://leetcode.com/problems/rotated-digits/

// 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;
    }
};

Java

class Solution {
    public int rotatedDigits(int N) {
        int count = 0;
        for (int i = 1; i <= N; i++) {
            String numStr = String.valueOf(i);
            if (numStr.indexOf('3') >= 0 || numStr.indexOf('4') >= 0 || numStr.indexOf('7') >= 0)
                continue;
            if (numStr.indexOf('2') >= 0 || numStr.indexOf('5') >= 0 || numStr.indexOf('6') >= 0 || numStr.indexOf('9') >= 0)
                count++;
        }
        return count;
    }
}

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