LeetCode //C - 788. Rotated Digits

788. Rotated Digits

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].
 

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:

• $1<=n<=10^4$

From: LeetCode
Link: 788. Rotated Digits

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Solution:

Ideas:

• Valid digits: {0, 1, 8} (same when rotated), {2, 5}, {6, 9} (change into each other).

• Invalid digits: {3, 4, 7} – make the number invalid after rotation.

• A good number must:

  • Not contain any invalid digits.
  • Contain at least one digit that changes to a different digit.

Code:

bool isGoodNumber(int num) {
    bool changed = false;

    while (num > 0) {
        int digit = num % 10;
        num /= 10;

        if (digit == 3 || digit == 4 || digit == 7)
            return false;

        if (digit == 2 || digit == 5 || digit == 6 || digit == 9)
            changed = true;
    }

    return changed;
}

int rotatedDigits(int n) {
    int count = 0;
    for (int i = 1; i <= n; i++) {
        if (isGoodNumber(i))
            count++;
    }
    return count;
}