Count Houses in a Circular Street II

class Solution {
    public int countOpenHouses(Street mainStreet, int maxSteps) {
        // Index of the first open door
        int firstOpenIdx = 0;

        // Distance from first open door to another open door
        int distance = 0;

        // Current walk index
        int stepIndex = 0;

        // Found the first open door
        boolean firstFound = false;

        while (stepIndex <= 2 * maxSteps) {
            // Check if the current door is open
            if (mainStreet.isDoorOpen()) {
                if (firstFound) {
                    distance = stepIndex - firstOpenIdx;
                    mainStreet.closeDoor();
                } else {
                    firstOpenIdx = stepIndex;
                    firstFound = true;
                }
            }

            mainStreet.moveRight();
            stepIndex++;
        }

        return distance;
    }
}

This solution involves solving the problem of counting the number of houses in a circular street using Java, focusing on the distance between the first two open houses within a specified range. The countOpenHouses method within the Solution class takes parameters mainStreet, a customized object representing the street (presumably with methods like isDoorOpen(), closeDoor(), and moveRight()), and maxSteps, an integer indicating the maximum steps that can be considered from the starting point.

Here’s a breakdown of how the functions within the code work:

• Initialize an index firstOpenIdx and a distance distance, both initially zero, to store the position of the first open house and the distance between the first two open houses found.
• Use stepIndex to track the number of steps taken, and a boolean flag firstFound to check the discovery of the first open house.
• Iterate while the condition stepIndex <= 2 * maxSteps holds true, effectively ensuring any house within twice the maximum steps can be considered to account for the circular nature of the street.
• Within this loop:
  • Check if the current door is open using the isDoorOpen() method.
  • If the first open house has been found (i.e., firstFound is true), compute the distance from firstOpenIdx to the current stepIndex and then close this currently open door.
  • If no open house has been located yet, log the current index as firstOpenIdx and set firstFound to true.
• After walking through the street up to the allowable range, return the calculated distance indicating the distance between the first two open houses detected.

This method offers a specific utility in scenarios such as security systems, neighborhood watch efficiency, or even urban planning simulation where understanding the distribution of certain characteristics (like open houses in this case) along a circular path is critical. The function employs straightforward iteration and condition checking, leveraging the properties and methods of the provided mainStreet object. Ensure to integrate error handling or consider cases where no open houses might be found within the given limits for broader application robustness.

class Solution:
    def countHouses(self, street: Optional['Street'], num: int) -> int:
        # Track distance to the initial open door found
        initial_open_door = 0

        # Distance from the current door to the initial open door
        distance_from_initial = 0

        # Current step in our traversal
        current_step = 0

        # Indicates if the initial open door has been found
        initial_found = False

        while current_step <= 2 * num:
            # Check if the current door is open
            if street.isDoorOpen(): 
                # If the initial door has been encountered before, record distance and close the door
                if initial_found:
                    distance_from_initial = current_step - initial_open_door
                    street.closeDoor()
                # Mark the first open door encountered
                else:
                    initial_open_door = current_step
                    initial_found = True

            # Move to the next house and increment the step counter
            street.moveRight()
            current_step += 1
        
        return distance_from_initial

In this solution, you're working with a problem of counting the number of houses between the first open door encountered and the next subsequent open door, given a circular street layout implemented in Python. The provided Python class Solution includes a method countHouses that takes two parameters: street, which is an instance of a Street class, and num, an integer representing the total number of houses on the street. This method efficiently determines the distance between two open doors on this circular street.

Detailed Steps:

1. Initialize several variables:

   • initial_open_door to store the position of the first open door encountered.
   • distance_from_initial to store the distance to the next open door encountered after the initial.
   • current_step to keep track of the current position on the street during traversal.
   • initial_found as a boolean to note when the first open door has been found.

2. Implement a while loop that will continue until current_step exceeds twice the value of num to ensure a complete scanning of the street considering its circular nature.

3. Inside the loop:

   • Check if the current door is open using street.isDoorOpen().
   • If an open door is found and initial_found is True, update distance_from_initial to the difference between current_step and initial_open_door and then invoke street.closeDoor().
   • If it's the first open door found (i.e., initial_found is False), update initial_open_door with current_step, and set initial_found to True.

4. Move to the next house using street.moveRight() and increment current_step by 1.

5. The function returns distance_from_initial, providing the distance between the first and the next open door encountered in the traversal.

This solution effectively leverages a single traversal to compute the desired distance, considering the circular nature of the street by allowing a loop around it up to twice the total number of doors. This approach ensures that the method will handle edge cases such as when the first open door is towards the very end of the list and the next one is towards the beginning.