How do you solve Count Number of Rectangles Containing Each Point on LeetCode?

Count Number of Rectangles Containing Each Point (LeetCode #2250) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(1) space complexity. Key insight: Height frequency can be used to quickly count rectangles for points.

What is the time complexity of Count Number of Rectangles Containing Each Point?

The optimal solution for Count Number of Rectangles Containing Each Point runs in O(n + m) time and uses O(1) space. The time complexity is O(n + m) because we process all rectangles once and then all points once, making it linear with respect to the input sizes.

What is the brute force solution for Count Number of Rectangles Containing Each Point?

The brute force approach for Count Number of Rectangles Containing Each Point is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each point against all rectangles to see if the point lies within any rectangle. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Count Number of Rectangles Containing Each Point?

Count Number of Rectangles Containing Each Point uses the following concepts: Array, Hash Table, Binary Search, Binary Indexed Tree, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Number of Rectangles Containing Each Point?

Clarify the problem constraints and edge cases before jumping into coding. Think about how you can optimize a brute force solution. Practice explaining your thought process as you code.

What are common mistakes when solving Count Number of Rectangles Containing Each Point?

Not considering edge cases where points lie exactly on rectangle boundaries. Overlooking the constraints on height and length.

#2250

Count Number of Rectangles Containing Each Point

Medium

ArrayHash TableBinary SearchBinary Indexed TreeSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n + m)Space O(1)

We can optimize the solution by sorting rectangles by their lengths and using a frequency array for heights to quickly count how many rectangles contain a given point's height.

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Algorithm

5 steps

1Step 1: Create a frequency array for heights (0 to 100) initialized to zero.
2Step 2: Sort rectangles by their lengths.
3Step 3: For each rectangle, update the frequency array for heights up to the rectangle's height.
4Step 4: For each point, count how many rectangles can contain it using the frequency array.
5Step 5: Return the result list.

solution.py9 lines

1def countRectangles(rectangles, points):
2    height_count = [0] * 101
3    for l, h in rectangles:
4        for i in range(h + 1):
5            height_count[i] += 1
6    result = []
7    for px, py in points:
8        result.append(height_count[py] if px <= 100 else 0)
9    return result

ℹ

Complexity note: The time complexity is O(n + m) because we process all rectangles once and then all points once, making it linear with respect to the input sizes.

1Height frequency can be used to quickly count rectangles for points.
2Sorting rectangles by length helps in efficiently managing point checks.

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