How do you solve K-th Nearest Obstacle Queries on LeetCode?

K-th Nearest Obstacle Queries (LeetCode #3275) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log k) time complexity and O(k) space complexity. Key insight: Using a max heap allows us to efficiently track the k smallest distances without sorting the entire list.

What is the brute force solution for K-th Nearest Obstacle Queries?

The brute force approach for K-th Nearest Obstacle Queries is Brute Force, with O(n²) time complexity and O(n) space complexity. In the brute force approach, we will calculate the distance of each obstacle from the origin after each query and sort these distances to find the k-th nearest obstacle. This is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n log k).

What algorithm or data structure is used to solve K-th Nearest Obstacle Queries?

K-th Nearest Obstacle Queries uses the following concepts: Array, Heap (Priority Queue). The recommended patterns to study are: Heap (Priority Queue), Sorting.

What are the interview tips for K-th Nearest Obstacle Queries?

Always consider edge cases, such as when there are fewer than k obstacles. Explain your thought process clearly; this shows your understanding of the problem. Practice implementing both brute force and optimal solutions to solidify your understanding.

What are common mistakes when solving K-th Nearest Obstacle Queries?

Forgetting to check if there are enough obstacles before accessing the k-th element. Not using a heap structure, leading to inefficient solutions.

#3275

K-th Nearest Obstacle Queries

Medium

ArrayHeap (Priority Queue)Heap (Priority Queue)Sorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log k)
Space	O(n)	O(k)

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Intuition

Time O(n log k)Space O(k)

In the optimal approach, we use a max heap to maintain the k nearest obstacles efficiently. This allows us to quickly find the k-th nearest obstacle without sorting the entire list each time.

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Algorithm

4 steps

1Step 1: Initialize a max heap to store the k smallest distances.
2Step 2: For each query, compute the distance from the origin and add it to the heap.
3Step 3: If the heap exceeds size k, remove the largest element (the root of the max heap).
4Step 4: After processing each query, if the heap has fewer than k elements, return -1; otherwise, return the root of the heap (the k-th nearest obstacle).

solution.py14 lines

1# Full working Python code
2import heapq
3from typing import List
4
5def kNearestObstacles(queries: List[List[int]], k: int) -> List[int]:
6    max_heap = []
7    results = []
8    for x, y in queries:
9        distance = abs(x) + abs(y)
10        heapq.heappush(max_heap, -distance)
11        if len(max_heap) > k:
12            heapq.heappop(max_heap)
13        results.append(-1 if len(max_heap) < k else -max_heap[0])
14    return results

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Complexity note: The time complexity is O(n log k) because we perform log k operations for each of the n queries to maintain the max heap. The space complexity is O(k) for storing the k nearest distances in the heap.

1Using a max heap allows us to efficiently track the k smallest distances without sorting the entire list.
2The k-th nearest obstacle can only be determined once we have at least k obstacles.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.