788. Rotated Digits LeetCode Solution

In this guide, you will get 788. Rotated Digits LeetCode Solution with the best time and space complexity. The solution to Rotated Digits problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Rotated Digits solution in C++
Rotated Digits solution in Java
Rotated Digits solution in Python
Additional Resources

788. Rotated Digits LeetCode Solution image

Problem Statement of Rotated Digits

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 <= 104

Complexity Analysis

Time Complexity: O(n\log n)
Space Complexity: O(1)

788. Rotated Digits LeetCode Solution in C++

class Solution {
 public:
  int rotatedDigits(int n) {
    int ans = 0;

    for (int i = 1; i <= n; ++i)
      if (isGoodNumber(i))
        ++ans;

    return ans;
  }

 private:
  bool isGoodNumber(int i) {
    bool isRotated = false;

    for (const char c : to_string(i)) {
      if (c == '0' || c == '1' || c == '8')
        continue;
      if (c == '2' || c == '5' || c == '6' || c == '9')
        isRotated = true;
      else
        return false;
    }

    return isRotated;
  }
};
/* code provided by PROGIEZ */

788. Rotated Digits LeetCode Solution in Java

class Solution {
  public int rotatedDigits(int n) {
    int ans = 0;

    for (int i = 1; i <= n; ++i)
      if (isGoodNumber(i))
        ++ans;

    return ans;
  }

  private boolean isGoodNumber(int i) {
    boolean isRotated = false;

    for (final char c : String.valueOf(i).toCharArray()) {
      if (c == '0' || c == '1' || c == '8')
        continue;
      if (c == '2' || c == '5' || c == '6' || c == '9')
        isRotated = true;
      else
        return false;
    }

    return isRotated;
  }
}
// code provided by PROGIEZ

788. Rotated Digits LeetCode Solution in Python

class Solution:
  def rotatedDigits(self, n: int) -> int:
    def isGoodNumber(i: int) -> bool:
      isRotated = False

      for c in str(i):
        if c == '0' or c == '1' or c == '8':
          continue
        if c == '2' or c == '5' or c == '6' or c == '9':
          isRotated = True
        else:
          return False

      return isRotated

    return sum(isGoodNumber(i) for i in range(1, n + 1))
 # code by PROGIEZ