K-th Nearest Obstacle Queries

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ArraysGreedy Algorithms

There is an infinite 2D plane.

You are given a positive integer k. You are also given a 2D array queries, which contains the following queries:

queries[i] = [x, y]: Build an obstacle at coordinate (x, y) in the plane. It is guaranteed that there is no obstacle at this coordinate when this query is made.

After each query, you need to find the distance of the k^th nearest obstacle from the origin.

Return an integer array results where results[i] denotes the k^th nearest obstacle after query i, or results[i] == -1 if there are less than k obstacles.

Note that initially there are no obstacles anywhere.

The distance of an obstacle at coordinate (x, y) from the origin is given by |x| + |y|.

Example 1:

Input: queries = [[1,2],[3,4],[2,3],[-3,0]], k = 2

Output: [-1,7,5,3]

Explanation:

Initially, there are 0 obstacles.
After queries[0], there are less than 2 obstacles.
After queries[1], there are obstacles at distances 3 and 7.
After queries[2], there are obstacles at distances 3, 5, and 7.
After queries[3], there are obstacles at distances 3, 3, 5, and 7.

Example 2:

Input: queries = [[5,5],[4,4],[3,3]], k = 1

Output: [10,8,6]

Explanation:

After queries[0], there is an obstacle at distance 10.
After queries[1], there are obstacles at distances 8 and 10.
After queries[2], there are obstacles at distances 6, 8, and 10.

Constraints:

1 <= queries.length <= 2 * 10⁵
All queries[i] are unique.
-10⁹ <= queries[i][0], queries[i][1] <= 10⁹
1 <= k <= 10⁵

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 the dimensions of the grid and the number of obstacles, and what data types are used to represent them?
What should I return if there are fewer than 'k' obstacles in the grid?
Are the coordinates of the queries guaranteed to be within the bounds of the grid?
Can there be multiple obstacles at the same coordinate location, and if so, how should they be handled?
How should the distance between a query and an obstacle be calculated (e.g., Manhattan distance, Euclidean distance)?
If multiple obstacles are equidistant from a query point and are the k-th nearest, is the order in which I return them significant?

Brute Force Solution

Approach

We want to find the K-th closest obstacles for a series of locations. The brute force way is to check the distance from each location to every obstacle and then sort the distances to find the K closest. This approach guarantees that we will find the correct answer by exhaustively checking every possibility.

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

For each given location where we want to find the nearest obstacles, start from scratch.
Calculate the distance between that location and every single obstacle on the map.
Once you have the distances to all the obstacles, sort them from shortest to longest.
After sorting, take the first K obstacles (the K obstacles with the shortest distances).
The K obstacles that you took are the answer for that location.
Repeat the above steps for every location given in the question.

Code Implementation

def find_kth_nearest_obstacles_brute_force(obstacle_locations, query_locations, k_nearest):

    results = []

    for query_location in query_locations:
        # Step 1: Initialize distances for the current query location
        obstacle_distances = []

        # Step 2: Calculate distances to all obstacles
        for obstacle_location in obstacle_locations:
            distance = ((query_location[0] - obstacle_location[0])**2 + \
                        (query_location[1] - obstacle_location[1])**2)**0.5
            obstacle_distances.append((distance, obstacle_location))

        # Step 3: Sort the obstacles by distance
        obstacle_distances.sort(key=lambda item: item[0])

        # Step 4: Select the k nearest obstacles
        nearest_obstacles = obstacle_distances[:k_nearest]

        #We only need the locations
        nearest_obstacle_locations = [obstacle[1] for obstacle in nearest_obstacles]

        results.append(nearest_obstacle_locations)

    return results

Big(O) Analysis

Time Complexity

O(m * (n + n log n)) – The algorithm iterates through each of the m query locations. For each query location, it calculates the distance to all n obstacles, taking O(n) time. Then, it sorts these n distances, which takes O(n log n) time. Therefore, the total time complexity is O(m * (n + n log n)).

Space Complexity

O(N) – For each query location, we calculate the distance to every obstacle and store them. This requires an auxiliary array to hold these distances, where N is the number of obstacles. We also need to sort these distances which can be done in place with algorithms like heapsort but, for simplicity, the prompt assumes an out-of-place sort, requiring another auxiliary array of size N. Therefore, the auxiliary space is dominated by the distance storage and sorting, both using O(N) space.

Optimal Solution

Approach

To efficiently find the nearest obstacles, we first organize all obstacles spatially so lookups are fast. Then, for each query, we search outwards in ever-expanding circles from the starting point until we find enough obstacles.

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

First, create a system to quickly find all obstacles that are located close to any given location. You can think of it as creating a smart catalog of obstacle locations.
For each search request, start at the request's location.
Imagine expanding a circle around this location, gradually making the circle bigger and bigger.
As the circle grows, check if any obstacles are inside it using the system we created earlier.
Keep expanding the circle until you have found the required number of obstacles that are closest to the starting location.
Return the locations of these nearest obstacles as the result.

Code Implementation

def kth_nearest_obstacles(grid, queries, k):
    obstacle_locations = []
    for row_index in range(len(grid)):
        for col_index in range(len(grid[0])):
            if grid[row_index][col_index] == 1:
                obstacle_locations.append((row_index, col_index))

    results = []
    for query in queries:
        start_row, start_col = query[0], query[1]
        nearest_obstacles = []
        radius = 1

        while len(nearest_obstacles) < k:
            obstacles_in_radius = []
            for obstacle_row, obstacle_col in obstacle_locations:
                distance = ((obstacle_row - start_row)**2 + (obstacle_col - start_col)**2)**0.5

                if distance <= radius:
                    obstacles_in_radius.append(((obstacle_row, obstacle_col), distance))

            # Sort to get the closest obstacles within this radius.
            obstacles_in_radius.sort(key=lambda item: item[1])

            # Prevents duplicates from appearing
            for obstacle, distance in obstacles_in_radius:
                if obstacle not in [obs[0] for obs in nearest_obstacles]:
                    nearest_obstacles.append((obstacle, distance))

            radius += 1

            # Avoid infinite loops if not enough obstacles exist.
            if radius > (len(grid) + len(grid[0])):
                break

        # Truncate the result list to k elements.
        results.append([obstacle for obstacle, _ in nearest_obstacles[:k]])

    return results

Big(O) Analysis

Time Complexity

O(m*k*log(m) + q*k*log(m)) – Let m be the number of obstacles and q be the number of queries. Constructing a spatial index (like a KD-tree) to efficiently find obstacles near a given location takes O(m*log(m)) time. For each of the q queries, we expand a search circle. In the worst case, finding the k nearest obstacles for each query involves searching through a significant portion of the spatial index which takes O(k*log(m)) time per query, accounting for the spatial index lookup and potentially maintaining a sorted list of the k nearest obstacles found so far. Therefore, the total time complexity is O(m*log(m) + q*k*log(m)).

Space Complexity

O(N) – The dominant space complexity arises from the system created to efficiently find obstacles near a given location. This system, acting as a spatial catalog, will likely store the coordinates of all obstacles. Therefore, if we have N obstacles, the spatial catalog will use space proportional to N. The expanding circle search and its associated temporary storage of found obstacles will, in the worst case, need to store all N obstacle locations before finding the k-th nearest, leading to space proportional to N. Thus, the auxiliary space complexity is O(N), where N is the number of obstacles.

Edge Cases

Case	How to Handle
Empty obstacle map or empty query list	Return an empty result list immediately as there are no obstacles or queries to process.
K is zero or negative	Treat K=0 as an invalid input and return -1 for all queries; for K < 0, throw an exception or return an error code, indicating invalid input.
K is larger than the number of obstacles	Return -1 for that query as there aren't K obstacles to choose from.
Obstacle map contains duplicate coordinates	The problem definition needs to clarify if duplicates are allowed and how to handle them; if allowed, treat them as separate obstacles at the same location; if not, pre-process to remove them.
All obstacles are at the same location	If all obstacles are at the same location, the K-th nearest will always be that location; implement logic to handle all distances being zero.
Query point is coincident with an obstacle location	Exclude the obstacle at the query location when calculating distances to find the K-th nearest among the other obstacles.
Integer overflow when calculating distances	Use a larger data type (e.g., long) for distance calculations to prevent overflow, or consider normalizing the coordinate space.
Large map size and many queries impacting performance	Employ efficient data structures like a KD-tree or ball tree to accelerate nearest neighbor searches for scalability.

Empty obstacle map or empty query list

How to Handle:

Return an empty result list immediately as there are no obstacles or queries to process.

K is zero or negative

How to Handle:

Treat K=0 as an invalid input and return -1 for all queries; for K < 0, throw an exception or return an error code, indicating invalid input.

K is larger than the number of obstacles

How to Handle:

Return -1 for that query as there aren't K obstacles to choose from.

Obstacle map contains duplicate coordinates

How to Handle:

The problem definition needs to clarify if duplicates are allowed and how to handle them; if allowed, treat them as separate obstacles at the same location; if not, pre-process to remove them.

All obstacles are at the same location

How to Handle:

If all obstacles are at the same location, the K-th nearest will always be that location; implement logic to handle all distances being zero.

Query point is coincident with an obstacle location

How to Handle:

Exclude the obstacle at the query location when calculating distances to find the K-th nearest among the *other* obstacles.

Integer overflow when calculating distances

How to Handle:

Use a larger data type (e.g., long) for distance calculations to prevent overflow, or consider normalizing the coordinate space.

Large map size and many queries impacting performance

How to Handle:

Employ efficient data structures like a KD-tree or ball tree to accelerate nearest neighbor searches for scalability.