Count the Number of Houses at a Certain Distance I

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You are given three positive integers n, x, and y.

In a city, there exist houses numbered 1 to n connected by n streets. There is a street connecting the house numbered i with the house numbered i + 1 for all 1 <= i <= n - 1 . An additional street connects the house numbered x with the house numbered y.

For each k, such that 1 <= k <= n, you need to find the number of pairs of houses (house₁, house₂) such that the minimum number of streets that need to be traveled to reach house₂ from house₁ is k.

Return a 1-indexed array result of length n where result[k] represents the total number of pairs of houses such that the minimum streets required to reach one house from the other is k.

Note that x and y can be equal.

Example 1:

Input: n = 3, x = 1, y = 3
Output: [6,0,0]
Explanation: Let's look at each pair of houses:
- For the pair (1, 2), we can go from house 1 to house 2 directly.
- For the pair (2, 1), we can go from house 2 to house 1 directly.
- For the pair (1, 3), we can go from house 1 to house 3 directly.
- For the pair (3, 1), we can go from house 3 to house 1 directly.
- For the pair (2, 3), we can go from house 2 to house 3 directly.
- For the pair (3, 2), we can go from house 3 to house 2 directly.

Example 2:

Input: n = 5, x = 2, y = 4
Output: [10,8,2,0,0]
Explanation: For each distance k the pairs are:
- For k == 1, the pairs are (1, 2), (2, 1), (2, 3), (3, 2), (2, 4), (4, 2), (3, 4), (4, 3), (4, 5), and (5, 4).
- For k == 2, the pairs are (1, 3), (3, 1), (1, 4), (4, 1), (2, 5), (5, 2), (3, 5), and (5, 3).
- For k == 3, the pairs are (1, 5), and (5, 1).
- For k == 4 and k == 5, there are no pairs.

Example 3:

Input: n = 4, x = 1, y = 1
Output: [6,4,2,0]
Explanation: For each distance k the pairs are:
- For k == 1, the pairs are (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), and (4, 3).
- For k == 2, the pairs are (1, 3), (3, 1), (2, 4), and (4, 2).
- For k == 3, the pairs are (1, 4), and (4, 1).
- For k == 4, there are no pairs.

Constraints:

2 <= n <= 100
1 <= x, y <= 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:

What are the constraints on n, start_pos, end_pos, and k? Specifically, what are the maximum and minimum values for each?
Is it possible for start_pos and end_pos to be equal, and if so, how should that be handled?
Can k be zero, and what should the output be in that case?
Are start_pos and end_pos guaranteed to be within the range [1, n]?
Could you provide an example input and the expected output to ensure my understanding of the problem is correct?

Brute Force Solution

Approach

To count houses at a specific distance, the brute force approach checks every single house pair against all possible road configurations. It calculates the distance between houses for each configuration and counts how many pairs match the target distance. This is like manually measuring distances between every possible pair of houses in every street layout and counting the matches.

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

First, consider all possible connections or paths between every pair of houses.
For each of these connection possibilities, calculate the distance between a given pair of houses.
Compare this calculated distance with the specific distance we are looking for.
If the calculated distance matches the specific distance, increase a counter by one.
Repeat this process for all possible house pairs and all possible path combinations.
The final count will give the number of house pairs at the specific distance.

Code Implementation

def count_houses_at_distance_brute_force(house_locations, target_distance):
    number_of_houses = len(house_locations)
    count = 0

    for first_house_index in range(number_of_houses):
        # Iterate through each house.

        for second_house_index in range(first_house_index + 1, number_of_houses):
            # Avoid double-counting pairs; start from the next house.

            distance = abs(house_locations[first_house_index] - house_locations[second_house_index])

            if distance == target_distance:
                count += 1
                # Increment when houses are at the target distance.

    return count

Big(O) Analysis

Time Complexity

O(n^4) – The brute force approach iterates through all possible pairs of houses which contributes O(n^2) where n is the number of houses. For each house pair, it considers all possible road configurations. Since the problem description indicates checking 'all possible connections or paths' between each pair (essentially enumerating all possible graphs or subsets of edges for each pair), and assuming in the worst case that the number of possible road configurations is proportional to the number of houses (or n), and calculating the distance for each configuration adds another factor of O(n). Therefore the total time complexity becomes O(n^2 * n * n) which simplifies to O(n^4).

Space Complexity

O(1) – The brute force approach iterates through all possible house pairs and connection possibilities, but the provided description doesn't mention any explicit auxiliary data structures being created. The only variables used seem to be a counter and variables to store calculated distances, which require constant space. Therefore, the space complexity is independent of the number of houses (N) and the space used remains constant.

Optimal Solution

Approach

The key is to realize that the distances between houses follow a predictable pattern based on the positions of the houses and the ends of the street. We can directly calculate how many houses are at a particular distance from the ends of the street without checking every house individually.

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

First, understand the street layout and the positions of the houses and points.
Consider each house and calculate its distance from the start and end of the street.
For each distance we are interested in, count how many houses are exactly that distance from either the start or end of the street. Keep a separate counter for each distance.
Return the counts for each requested distance.

Code Implementation

def count_houses_at_distance(house_locations, special_house_locations, target_distance):
    houses_at_target_distance = set()

    # Iterate through each special house.
    for special_house_location in special_house_locations:
        # Calculate the location of houses at the target distance.
        house_location_left = special_house_location - target_distance
        house_location_right = special_house_location + target_distance

        # Check if the house exists and add it to the set.
        if house_location_left in house_locations:
            houses_at_target_distance.add(house_location_left)

        # Avoid double-counting if left and right are the same.
        if house_location_right in house_locations:
            houses_at_target_distance.add(house_location_right)

    # Return the total count of houses at the target distance.
    return len(houses_at_target_distance)

Big(O) Analysis

Time Complexity

O(n + k) – The algorithm iterates through each of the 'n' houses to calculate its distance from the start and end of the street. Then, for each of the 'k' requested distances, it iterates through all 'n' houses again to count those that match the target distance. Therefore, the overall time complexity is O(n + k*n) which can be simplified to O(n + k) assuming k is relatively small compared to n.

Space Complexity

O(K) – The algorithm maintains separate counters for each distance we are interested in, as stated in step 3. If we are interested in K different distances, then we require K counters to store the number of houses at each distance. Therefore, the auxiliary space is proportional to the number of distances, K, and the space complexity is O(K). The input size, N (number of houses), does not directly affect the auxiliary space used for the counters.

Edge Cases

Case	How to Handle
n is 0 or negative	Return an empty array since there are no houses.
k is negative	Since distance cannot be negative, return an array of zeros.
start_pos or end_pos are outside the range [1, n]	Adjust start_pos and end_pos to the valid range [1, n].
start_pos and end_pos are the same	The distance calculation should only consider this position once to prevent double-counting.
k is very large, larger than the maximum possible distance between houses.	The counts array will consist entirely of zeros in this scenario and will be efficiently calculated.
n is very large, leading to potential memory issues with the counts array.	Ensure memory allocation for the 'counts' array doesn't exceed available memory, consider a more memory-efficient representation if feasible, but it will still need to store n counts so consider the use case for such a large n.
start_pos and end_pos are far apart, and k is close to n, leading to many houses at distance k.	The solution iterates through all houses and efficiently calculates distances, handling this scenario without issues, although the runtime is proportional to n.
Integer overflow if using languages like C/C++ to calculate distance and n is extremely large.	Use a data type that can accommodate larger numbers (e.g., long long) to prevent integer overflow.

n is 0 or negative

How to Handle:

Return an empty array since there are no houses.

k is negative

How to Handle:

Since distance cannot be negative, return an array of zeros.

start_pos or end_pos are outside the range [1, n]

How to Handle:

Adjust start_pos and end_pos to the valid range [1, n].

start_pos and end_pos are the same

How to Handle:

The distance calculation should only consider this position once to prevent double-counting.

k is very large, larger than the maximum possible distance between houses.

How to Handle:

The counts array will consist entirely of zeros in this scenario and will be efficiently calculated.

n is very large, leading to potential memory issues with the counts array.

How to Handle:

Ensure memory allocation for the 'counts' array doesn't exceed available memory, consider a more memory-efficient representation if feasible, but it will still need to store n counts so consider the use case for such a large n.

start_pos and end_pos are far apart, and k is close to n, leading to many houses at distance k.

How to Handle:

The solution iterates through all houses and efficiently calculates distances, handling this scenario without issues, although the runtime is proportional to n.

Integer overflow if using languages like C/C++ to calculate distance and n is extremely large.

How to Handle:

Use a data type that can accommodate larger numbers (e.g., long long) to prevent integer overflow.