788. Rotated Digits

Problem

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

Solution (Java)

class Solution {
    public int rotatedDigits(int n) {
        int[] flag = new int[n + 1];
        flag[0] = 2;
        int[] indexesValueTwo = {1, 8};
        for (int value : indexesValueTwo) {
            if (n >= value) {
                flag[value] = 2;
            }
        }
        int[] indexesValueOne = {2, 5, 6, 9};
        for (int value : indexesValueOne) {
            if (n >= value) {
                flag[value] = 1;
            }
        }
        int rs = 0;
        for (int i = 1; i <= n; i++) {
            int residual = i % 10;
            if (flag[residual] != 0) {
                if ((residual == 1 || residual == 0 || residual == 8) && (flag[i / 10] == 2)) {
                    flag[i] = 2;
                    continue;
                }
                if (flag[i / 10] != 0) {
                    flag[i] = 1;
                    rs++;
                }
            }
        }
        return rs;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).