How do you solve Block Placement Queries on LeetCode?

Block Placement Queries (LeetCode #3161) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using sorted data structures allows for efficient nearest obstacle queries.

What is the brute force solution for Block Placement Queries?

The brute force approach for Block Placement Queries is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible position for placing a block of size `sz` in the range `[0, x]` while ensuring it does not overlap with any obstacles. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Block Placement Queries?

Think about how to efficiently manage dynamic data like obstacles. Consider using data structures that support fast insertions and queries. Practice explaining your thought process clearly, as it helps in interviews.

What are common mistakes when solving Block Placement Queries?

Not maintaining the order of obstacles, leading to incorrect checks. Overlooking edge cases where the block size equals the distance to the nearest obstacle.

#3161

Block Placement Queries

Hard

ArrayBinary SearchBinary Indexed TreeSegment TreeHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

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

The optimal solution uses a data structure to efficiently manage obstacles and quickly determine if a block can be placed. By maintaining a sorted list of obstacles, we can use binary search to find the nearest obstacle and check placement conditions in logarithmic time.

⚙️

Algorithm

3 steps

1Step 1: Initialize a sorted list to store obstacles.
2Step 2: For each query, if it's type 1, insert the obstacle in sorted order.
3Step 3: For each query of type 2, use binary search to find the nearest obstacle within the range and check if the block can fit.

solution.py16 lines

1from bisect import insort, bisect_right
2
3def blockPlacementQueries(queries):
4    obstacles = []
5    results = []
6    for query in queries:
7        if query[0] == 1:
8            insort(obstacles, query[1])
9        elif query[0] == 2:
10            x, sz = query[1], query[2]
11            pos = bisect_right(obstacles, x)
12            canPlace = True
13            if pos > 0 and obstacles[pos - 1] >= x - sz:
14                canPlace = False
15            results.append(canPlace)
16    return results

ℹ

Complexity note: The complexity is O(n log n) due to the insertion of obstacles into a sorted structure and the binary search for nearest obstacles, which is efficient for large inputs.

1Using sorted data structures allows for efficient nearest obstacle queries.
2Binary search can significantly reduce the time complexity of range queries.

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