LeetCode 788: Rotated Digits Solution
Master LeetCode problem 788 (Rotated Digits), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
788. Rotated Digits
Problem Explanation
Explanation
To solve this problem, we iterate through numbers from 1 to n and check if each digit in the number can be rotated to a different valid number. We can use a helper function to check if a number is good based on the rotating rules provided. If the number is good, we increment a counter. Finally, we return the count of good numbers found within the range [1, n].
- Time complexity: O(n * log(n)) where log(n) is the number of digits in n
- Space complexity: O(1)
Solution Code
class Solution {
public int rotatedDigits(int n) {
int count = 0;
for (int i = 1; i <= n; i++) {
if (isGood(i)) {
count++;
}
}
return count;
}
private boolean isGood(int num) {
boolean valid = false;
while (num > 0) {
int digit = num % 10;
if (digit == 3 || digit == 4 || digit == 7) {
return false;
}
if (digit == 2 || digit == 5 || digit == 6 || digit == 9) {
valid = true;
}
num /= 10;
}
return valid;
}
}Try It Yourself
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