Count Houses in a Circular Street

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There are n houses evenly placed along a circular street. The houses are numbered from 0 to n - 1. A ধনকuber driver is given the task to deliver a package to all of the houses, one package per house.

The ধনকuber driver starts at house number start and must deliver all of the packages in the order they appear in the array houses. More specifically, the driver must go to house houses[0] first, deliver the package, then go to house houses[1], and so on. Delivering all of the packages in the given order completes the task.

You are given the integer n, the integer start, and the array houses. You should return the minimum number of moves required to deliver all of the packages.

Example 1:

Input: n = 5, start = 0, houses = [1,3,4]
Output: 4
Explanation: 
- To reach house 1 from house 0: use the right direction, 1 move.
- To reach house 3 from house 1: use the right direction, 2 moves.
- To reach house 4 from house 3: use the right direction, 1 move.
Total moves: 1 + 2 + 1 = 4.

Example 2:

Input: n = 4, start = 0, houses = [2,0]
Output: 3
Explanation: 
- To reach house 2 from house 0: use the right direction, 2 moves.
- To reach house 0 from house 2: use the left direction, 2 moves.
Total moves: 2 + 1 = 3.

Example 3:

Input: n = 10, start = 2, houses = [0,3,8,3]
Output: 14
Explanation: 
- To reach house 0 from house 2: use the left direction, 2 moves.
- To reach house 3 from house 0: use the right direction, 3 moves.
- To reach house 8 from house 3: use the right direction, 5 moves.
- To reach house 3 from house 8: use the left direction, 5 moves.
Total moves: 2 + 3 + 5 + 4 = 14.

Constraints:

2 <= n <= 10⁵
0 <= start < n
1 <= houses.length <= 10⁵
0 <= houses[i] < n

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can you please describe how the houses are represented as an input? Is it an array of house values or another data structure, and what does each value signify?
What are the possible value ranges for the house values, and are negative values, zero, or null values possible?
Is there a defined starting point in the circular street, or can I assume any house can be the starting point for counting?
If there are no houses or an invalid input, what should I return?
Are there any constraints on the number of houses that can be on the street? What is the maximum number of houses that would be tested?

Brute Force Solution

Approach

The brute force approach to counting houses exhaustively explores every possible combination of house assignments in the circular street. It checks each combination to see if it meets a certain condition, like whether houses are similar to each other. This continues until all possible combinations have been evaluated.

Here's how the algorithm would work step-by-step:

Start by considering the first house; we'll try assigning it to be of a specific type.
Next, for the second house, we will try assigning every possible type to it.
Continue this process, trying every type assignment for each house in the street, one at a time.
After assigning a type to every single house, check if the assignment of all houses follows the rules (e.g., no similar houses next to each other).
If the houses follow the rules, then increase our count of valid street arrangements.
Repeat the entire process, starting back at the first house, trying all possible types for each house until we have exhausted all possibilities for all houses.
Once done exploring every single house type combination, the final count is the total number of valid house arrangements following all rules.

Code Implementation

def count_houses_brute_force(number_of_houses, house_types, are_similar):
    arrangement_count = 0

    def is_valid_arrangement(arrangement):
        # Check for similarity between adjacent houses.
        for i in range(number_of_houses):
            if are_similar(arrangement[i], arrangement[(i + 1) % number_of_houses]):
                return False
        return True

    def generate_arrangements(current_arrangement):
        nonlocal arrangement_count

        if len(current_arrangement) == number_of_houses:
            # Evaluate the completed arrangement.
            if is_valid_arrangement(current_arrangement):
                arrangement_count += 1
            return

        # Try assigning each type to the next house.
        for house_type in house_types:

            # Recursively build the house arrangement
            generate_arrangements(current_arrangement + [house_type])

    # Start the generation process with an empty arrangement.
    generate_arrangements([])

    return arrangement_count

Big(O) Analysis

Time Complexity

O(k^n) – The brute force approach explores all possible combinations of house types. If there are n houses and k possible types for each house, the algorithm essentially generates all possible strings of length n where each character can be one of k options. This results in k^n possible combinations to check. For each combination, the algorithm checks if it is valid. Therefore, the time complexity is O(k^n).

Space Complexity

O(N) – The brute force approach described utilizes recursion implicitly as it explores each possible house assignment. Specifically, the recursion's call stack depth can reach a maximum of N, where N is the number of houses in the circular street, because each recursive call corresponds to assigning a type to one house. Each stack frame stores local variables and the return address, leading to a space complexity proportional to the depth of the recursion. Therefore, the auxiliary space required is O(N).

Optimal Solution

Approach

The key idea is to use math to figure out how many houses we can safely count without actually walking around the entire circle. We can establish relationships between counted and uncounted houses to find the answer efficiently.

Here's how the algorithm would work step-by-step:

First, think about what happens if we count some houses and skip others. We're told some houses are visible, and some are not.
Realize that when you skip a house, it impacts how many of the houses after it can potentially be visible. If you skip too many houses in a row, it affects how many total houses there are.
Try to imagine the simplest situation: what if you skip just one house? How many houses can you count after that?
Now, see how the number of houses you count and the number you skip relate to the total number of houses. The total number has to add up.
By figuring out a minimum number of houses that MUST be skipped and comparing it to what is skipped, we can determine the number of skipped houses.
From that known skipped number we can work backward to figure out the number of visible houses.

Code Implementation

def count_houses(visible_houses: int, skipped_houses: int) -> int:
    # Minimum skipped is 1, otherwise all houses are visible.
    if skipped_houses == 0:
        return visible_houses

    minimum_skipped_houses = 1

    # Determine total houses with the equation given number of visible houses
    # and the minimum number of skipped houses.
    total_houses_minimum = visible_houses + (visible_houses - 1) * minimum_skipped_houses

    # We need to calculate how many more houses are skipped beyond the minimum.
    houses_skipped_beyond_minimum = skipped_houses - minimum_skipped_houses

    # Each house skipped beyond the minimum adds one to the total count.
    total_houses = total_houses_minimum + houses_skipped_beyond_minimum

    return total_houses

Big(O) Analysis

Time Complexity

O(n) – The algorithm's time complexity is dominated by the initial house counting and visibility check. This process iterates through the houses at most once, relating counted and skipped houses to the total number of houses (n). Determining the number of skipped houses from this relationship requires a constant number of operations. Consequently, the overall time complexity scales linearly with the number of houses, resulting in O(n).

Space Complexity

O(1) – The plain English explanation focuses on mathematical relationships and deductions rather than data storage. It speaks of counting, skipping, and comparing numbers, which typically involve constant space variables for intermediate calculations and counts. No data structures that grow with the input size, such as arrays or hash maps, are mentioned. Therefore, the auxiliary space complexity is constant, independent of the number of houses (N).

Edge Cases

Case	How to Handle
Null or empty street array	Return 0 immediately as no houses exist.
Street with only one house	Return 1 as only one house is present in the street.
Street with two houses	Return 2 as two houses are present in the street.
All houses have the same value	The number of houses is equal to the length of street array.
Negative house values	The solution correctly counts houses irrespective of negative values if it is correctly defined based on the problem description.
Maximum integer value for the number of houses	Ensure that the data type used to store the house count can handle large values to avoid overflow.
Very large street array size that could lead to memory issues	Check and document memory usage constraints to avoid memory issues in extreme scale cases.
Integer overflow when calculating sum of houses or other calculations	Use appropriate data types (e.g., long) to prevent integer overflow during calculations.

Null or empty street array

How to Handle:

Return 0 immediately as no houses exist.

Street with only one house

How to Handle:

Return 1 as only one house is present in the street.

Street with two houses

How to Handle:

Return 2 as two houses are present in the street.

All houses have the same value

How to Handle:

The number of houses is equal to the length of street array.

Negative house values

How to Handle:

The solution correctly counts houses irrespective of negative values if it is correctly defined based on the problem description.

Maximum integer value for the number of houses

How to Handle:

Ensure that the data type used to store the house count can handle large values to avoid overflow.

Very large street array size that could lead to memory issues

How to Handle:

Check and document memory usage constraints to avoid memory issues in extreme scale cases.

Integer overflow when calculating sum of houses or other calculations

How to Handle:

Use appropriate data types (e.g., long) to prevent integer overflow during calculations.