788. Rotated Digits

Posted 2021-10-28 ruruozhenhao

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].

 

Approach #1: DP. [Java]

class Solution {
    public int rotatedDigits(int N) {
        int[] dp = new int[N+1];
        int count = 0;
        for (int i = 0; i <= N; ++i) {
            if (i < 10) {
                if (i == 0 || i == 1 || i == 8) dp[i] = 1;
                if (i == 2 || i == 5 || i == 6 || i == 9) {
                    dp[i] = 2;
                    count++;
                }
            } else {
                int a = dp[i/10], b = dp[i%10];
                if (a == 1 && b == 1) dp[i] = 1;
                else if (a >= 1 && b >= 1) {
                    dp[i] = 2;
                    count++;
                }
            }
        }
        return count;
    }
}

　　

Analysis:

dp[0] : invalid number

dp[1]: valid and same number

dp[2]: valid and difference number

 

Reference:

https://leetcode.com/problems/rotated-digits/discuss/117975/Java-dp-solution-9ms