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

Maximum Area Rectangle With Point Constraints I (LeetCode #3380) 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: Using a set allows for quick checks of point existence.

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

The optimal solution for Maximum Area Rectangle With Point Constraints I runs in O(n²) time and uses O(n) space. The time complexity remains O(n²) due to the nested loops, but we use O(n) space for the set to allow for quick point existence checks.

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

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

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

Always consider edge cases, such as points lying on the border. Explain your thought process clearly while coding. Practice identifying patterns in problems to improve problem-solving speed.

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

Not checking for points inside the rectangle. Assuming all combinations of four points will yield valid rectangles.

#3380

Maximum Area Rectangle With Point Constraints I

Medium

ArrayMathBinary Indexed TreeSegment TreeGeometrySortingEnumerationHash 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)

By fixing two points as opposite corners of a rectangle, we can efficiently find the other two corners and check for point constraints using a set for quick lookups.

⚙️

Algorithm

6 steps

1Step 1: Store all points in a set for O(1) lookups.
2Step 2: Iterate through all pairs of points to consider them as opposite corners of a rectangle.
3Step 3: Calculate the other two corners based on the current pair.
4Step 4: Check if these two corners exist in the set.
5Step 5: Check if any point lies inside or on the border of the rectangle using the set.
6Step 6: Calculate the area if valid and update the maximum area found.

solution.py14 lines

1def maxArea(points):
2    point_set = set(map(tuple, points))
3    max_area = -1
4    n = len(points)
5    for i in range(n):
6        for j in range(i + 1, n):
7            x1, y1 = points[i]
8            x2, y2 = points[j]
9            if x1 != x2 and y1 != y2:
10                if (x1, y2) in point_set and (x2, y1) in point_set:
11                    area = abs(x1 - x2) * abs(y1 - y2)
12                    if all(not (min(x1, x2) <= px <= max(x1, x2) and min(y1, y2) <= py <= max(y1, y2)) for px, py in points if (px, py) not in {(x1, y1), (x2, y2)}):
13                        max_area = max(max_area, area)
14    return max_area

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we use O(n) space for the set to allow for quick point existence checks.

1Using a set allows for quick checks of point existence.
2Fixing two points reduces the complexity of finding the other corners.

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