[LeetCode] Problem 788 - Rotated Digits

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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].

Code

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private int INVALID = (1 << 3) | (1 << 4) | (1 << 7);
private int VALID = (1 << 2) | (1 << 5) | (1 << 6) | (1 << 9);

public int rotatedDigits(int N) {
    int result = 0;

    for (int i = 1; i <= N; i++) {
        if (isValid(i))
            result++;
    }

    return result;
}

private boolean isValid(int n) {
    boolean result = false;

    while (n > 0) {
        int bit = 1 << (n % 10);

        if ((bit & INVALID) != 0)
            return false;
        else if ((bit & VALID) != 0)
            result = true;

        n /= 10;
    }

    return result;
}