788. Rotated Digits
Tags: ‘String’, ‘Dynamic Programming’
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].
Solution using Dynamic Programming
O(n) time, O(n) space
public int rotatedDigits(int N) {
/*
dp[i] = x
x = 0: invalid
x = 1: same;
x = 2: valid
State Transition:
a = num/10 previous number
b = num%10 last digit to append
if a or b = 0, reuslt = 0
if both a and b = 1, result = 1
if at least one = 2, result = 2
*/
int[] dp = new int[N + 1]; // default 0
int count = 0;
for (int i = 0; i <= N; i++) {
if (i < 10) {
if (i == 0 || i == 1 || i == 8) {
dp[i] = 1;
} else if (i == 6 || i == 9 || i ==2 || i == 5) {
dp[i] = 2;
count++;
}
} else {
int a = dp[i/10], b = dp[i%10];
if (a == 0 || b == 0) {
dp[i] = 0;
} else if (a == 1 && b == 1) {
dp[i] = 1;
} else if (a == 2 || b == 2) {
dp[i] = 2;
count++;
}
}
}
return count;
}