How do you solve Maximum Area Rectangle With Point Constraints II on LeetCode?

Maximum Area Rectangle With Point Constraints II (LeetCode #3382) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Sorting helps in efficiently checking valid rectangles.

What is the time complexity of Maximum Area Rectangle With Point Constraints II?

The optimal solution for Maximum Area Rectangle With Point Constraints II runs in O(n²) time and uses O(n) space. Sorting takes O(n log n) and checking pairs takes O(n²), but using a set allows for quick lookups.

What is the brute force solution for Maximum Area Rectangle With Point Constraints II?

The brute force approach for Maximum Area Rectangle With Point Constraints II is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all combinations of four points to see if they form a valid rectangle. This method is straightforward but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Maximum Area Rectangle With Point Constraints II?

Maximum Area Rectangle With Point Constraints II uses the following concepts: Array, Math, Binary Indexed Tree, Segment Tree, Geometry, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Area Rectangle With Point Constraints II?

Explain your thought process clearly. Discuss edge cases, like all points being collinear. Practice similar geometry problems to build intuition.

What are common mistakes when solving Maximum Area Rectangle With Point Constraints II?

Not checking for points on the border. Ignoring duplicate x-coordinates.

#3382

Maximum Area Rectangle With Point Constraints II

Hard

ArrayMathBinary Indexed TreeSegment TreeGeometrySortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

Sort points by x-coordinates and then by y-coordinates. Use a set to track points and check for valid rectangles efficiently.

⚙️

Algorithm

3 steps

1Step 1: Sort points by x-coordinates, then by y-coordinates.
2Step 2: Use a nested loop to select pairs of y-coordinates for each unique x-coordinate.
3Step 3: Check if the rectangle formed by these pairs has no other points inside or on the border using a set.

solution.py12 lines

1def maxArea(xCoord, yCoord):
2    points = sorted(zip(xCoord, yCoord))
3    point_set = set(points)
4    max_area = -1
5    for i in range(len(points)):
6        for j in range(i + 1, len(points)):
7            if points[i][0] != points[j][0]:
8                y1, y2 = points[i][1], points[j][1]
9                if not any((x, y1) in point_set or (x, y2) in point_set for x in range(points[i][0] + 1, points[j][0])):
10                    area = abs(points[i][0] - points[j][0]) * abs(y1 - y2)
11                    max_area = max(max_area, area)
12    return max_area

ℹ

Complexity note: Sorting takes O(n log n) and checking pairs takes O(n²), but using a set allows for quick lookups.

1Sorting helps in efficiently checking valid rectangles.
2Using a set allows for O(1) lookups to check for points inside.

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