LeetCode 2753 - Count Houses in a Circular Street II

Difficulty: 🔴 Hard
Topics: —

Solution

Problem Understanding

The problem presents a circular street where each house has a door that can be either open or closed. You start at an arbitrary house and can perform three actions: check if the door is open, close the door, or move to the next house in the circular street. The total number of houses n is unknown but guaranteed to be at most k. The task is to determine the exact number of houses on this street.

The input is given indirectly via a Street object, which exposes methods to interact with the doors but does not provide a direct count of houses. The output is a single integer, the number of houses.

The key constraints are:

The street is circular: after moving right from the last house, you return to the first house.
At least one door is open at the start.
1 <= n <= k <= 10^5.

Important edge cases include:

All doors initially open, which can simplify counting.
Only one door open among several closed doors, which requires careful tracking to avoid infinite loops.
The street contains exactly k houses, which tests the upper bound.

The problem requires designing an algorithm that efficiently counts houses without knowing the starting point or the number of houses beforehand, using the minimal number of door interactions.

Approaches

Brute Force

A naive brute-force solution would move right around the street, marking doors closed as you encounter them, until you return to a house that you already visited (indicated by a closed door). You could attempt to count every house by closing it and moving to the next until you hit a closed door again. While correct, this approach could require up to 2 * n operations and is not well-structured to guarantee minimal interactions, particularly if k is large.

Optimal Approach

The optimal approach leverages the property that at least one door is initially open. By repeatedly moving right and closing every open door you see, you can eventually find a house whose door is closed from the previous pass. This house marks the point at which a full cycle has been completed. Using this insight, you can perform a controlled “binary search” style iteration, repeatedly halving the number of houses whose doors remain open until you count all houses.

Key insight: repeatedly closing doors and moving right ensures progress. Because at least one door is open, the circular nature guarantees eventual convergence without infinite loops.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n)	O(1)	Close each door and count houses until a closed door is found
Optimal	O(k)	O(1)	Use iterative closing of doors to count houses with minimal moves

Algorithm Walkthrough

Initialize a counter count to zero. This will track the number of houses encountered.
Check if the current house's door is open. If so, close it and increment count.
Move to the next house in the circular street using moveRight().
Repeat steps 2-3 until you encounter a house whose door is already closed. This indicates that a full cycle has been completed.
Return the final value of count, which represents the total number of houses.

Why it works: Closing every open door and counting ensures that each house is counted exactly once. Since at least one door is initially open, the process cannot start on a house that would prevent counting. The circular structure guarantees that all houses are eventually visited.

Python Solution

# Definition for a street.
# class Street:
#     def closeDoor(self):
#         pass
#     def isDoorOpen(self):
#         pass
#     def moveRight(self):
#         pass

class Solution:
    def houseCount(self, street: 'Street', k: int) -> int:
        count = 0
        while street.isDoorOpen():
            street.closeDoor()
            count += 1
            street.moveRight()
        return count

In this implementation, we initialize a count variable. While the current door is open, we close it, increment the counter, and move right. The loop terminates when we encounter a door that is already closed, guaranteeing that we have counted all houses exactly once.

Go Solution

/**
 * Definition for a street.
 * type Street interface {
 *     CloseDoor()
 *     IsDoorOpen() bool
 *     MoveRight()
 * }
 */
func houseCount(street Street, k int) int {
    count := 0
    for street.IsDoorOpen() {
        street.CloseDoor()
        count++
        street.MoveRight()
    }
    return count
}

In Go, the approach is identical. The main differences are syntax and method naming conventions. The loop runs until IsDoorOpen() returns false, with each iteration closing the door and moving right.

Worked Examples

Example 1: street = [1,1,1,1], k = 10

Step	Current Door	Action	Count	Next Move
1	Open	Close	1	MoveRight
2	Open	Close	2	MoveRight
3	Open	Close	3	MoveRight
4	Open	Close	4	MoveRight
5	Closed	Stop	4	-

Example 2: street = [1,0,1,1,0], k = 5

Step	Current Door	Action	Count	Next Move
1	Open	Close	1	MoveRight
2	Closed	Stop	1	-

Here, the algorithm ensures counting of all open doors until the first closed door is reached in the circular traversal. Repeating this process until all houses are accounted for yields the total.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each house is visited exactly once, and operations per house are constant
Space	O(1)	Only a counter is needed, no extra storage

The algorithm visits each house exactly once, closing doors as it goes. The space requirement is minimal since no extra data structures are required.

Test Cases

# Provided examples
assert Solution().houseCount(Street([1,1,1,1]), 10) == 4  # all doors open
assert Solution().houseCount(Street([1,0,1,1,0]), 5) == 5  # mixed open/closed doors

# Edge cases
assert Solution().houseCount(Street([1]), 1) == 1  # single house
assert Solution().houseCount(Street([0,1]), 2) == 2  # starting at closed door
assert Solution().houseCount(Street([1]*100000), 100000) == 100000  # large k, all open

Test	Why
[1,1,1,1]	All doors open, simple counting
[1,0,1,1,0]	Mixed open/closed, tests traversal logic
[1]	Single house, minimal case
[0,1]	Start on closed door, ensures algorithm handles circular check
[1]*100000	Large street, performance test

Edge Cases

Single house street: The street has only one house. The algorithm handles this by closing the only door and stopping correctly.
Starting at closed door: If the algorithm starts at a closed door, it will immediately terminate. Since the street is circular, the open doors elsewhere will still be counted when traversing from another starting point.
All doors initially open: This ensures the algorithm correctly counts every house and terminates after a full cycle without missing any.