788. Rotated Digits

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].

Smart DP representing three states source

public int rotatedDigits(int N) {
    int[] dp = new int[1+N];
    // 0 -> invalid;
    // 1 -> retate to itself;
    // 2 -> rotate to a different number

    int result = 0;
    for(int i = 0; i <= N; i++) {
        if(i <= 9) {
            if(i == 0 || i == 1 || i == 8) dp[i] = 1;
            else if(i == 2 || i == 5 || i == 6 || i == 9) {
                dp[i] = 2;
                result++;
            }
        } else {
            int prefix = dp[i / 10], lastDigit = dp[i % 10];
            if(prefix == 0 || lastDigit == 0) dp[i] = 0;
            else if(prefix == 1 && lastDigit == 1)  dp[i] = 1;
            else {
                dp[i] = 2;
                result++;
            }
        }
    }
    return result;
}

Brutal force to generate all candidates

class Solution {
    int result = 0;
    int[] arr = {0, 1, 8, 2, 5, 6, 9};
    int N;
    HashSet<Integer> set2569 = new HashSet<>(Arrays.asList(2,5,6,9));
    public int rotatedDigits(int N) {
        this.N = N;
        for(int i = 1; i < arr.length; i++) {
            count(arr[i], set2569.contains(arr[i]));
        }
        return result;
    }

    public void count(int curr, boolean has2569) {
        if(curr > N) return;
        if(has2569) result++;
        for(int i : arr) {
            count(curr * 10 + i, has2569 || set2569.contains(i));
        }
    }
}