Maximum Area of Longest Diagonal Rectangle

Easy

Atlassian

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Topics:

Arrays

You are given a 2D 0-indexed integer array dimensions.

For all indices i, 0 <= i < dimensions.length, dimensions[i][0] represents the length and dimensions[i][1] represents the width of the rectangle i.

Return the area of the rectangle having the longest diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the maximum area.

Example 1:

Input: dimensions = [[9,3],[8,6]]
Output: 48
Explanation: 
For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487.
For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.
So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48.

Example 2:

Input: dimensions = [[3,4],[4,3]]
Output: 12
Explanation: Length of diagonal is the same for both which is 5, so maximum area = 12.

Constraints:

1 <= dimensions.length <= 100
dimensions[i].length == 2
1 <= dimensions[i][0], dimensions[i][1] <= 100

Maximum Area of Longest Diagonal Rectangle

Easy

Atlassian

7 views

Topics:

Arrays

You are given a 2D 0-indexed integer array dimensions.

For all indices i, 0 <= i < dimensions.length, dimensions[i][0] represents the length and dimensions[i][1] represents the width of the rectangle i.

Return the area of the rectangle having the longest diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the maximum area.

Example 1:

Input: dimensions = [[9,3],[8,6]]
Output: 48
Explanation: 
For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487.
For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.
So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48.

Example 2:

Input: dimensions = [[3,4],[4,3]]
Output: 12
Explanation: Length of diagonal is the same for both which is 5, so maximum area = 12.

Constraints:

1 <= dimensions.length <= 100
dimensions[i].length == 2
1 <= dimensions[i][0], dimensions[i][1] <= 100

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the length and width of each rectangle? Can they be zero or negative?
What is the expected return value if the input array `dimensions` is empty or null?
If multiple rectangles have the same longest diagonal, should I return the area of the first one encountered, or the rectangle with the maximum area among those with the longest diagonal?
What is the data type of the length and width values in the `dimensions` array (e.g., integers, floating-point numbers)? If floating-point numbers, what precision is required?
Are there any upper bounds on the length and width, and should I be concerned about potential integer overflow when calculating the diagonal or the area?

Brute Force Solution

Approach

The brute force method involves checking every possible combination of rectangles that can be formed from the given points. We'll calculate the length of the diagonal for each rectangle and then its area, ultimately finding the maximum area among those with the longest diagonal length.

Here's how the algorithm would work step-by-step:

Consider every possible pair of points from the input. Each pair could potentially form the diagonal of a rectangle.
For each pair of points (potential diagonal), examine all other points to see if any two of these remaining points could form the other two corners of a rectangle with that diagonal.
To confirm we have a rectangle, ensure that the sides formed by the points are parallel to the x and y axes.
If a rectangle is found, calculate the length of its diagonal.
Identify the maximum diagonal length found so far across all rectangles.
For each rectangle, if its diagonal length equals the maximum diagonal length, calculate the area of the rectangle.
Keep track of the maximum area amongst all rectangles with the maximum diagonal length.
After checking all possible rectangles, report the maximum area found.

Code Implementation

def maximum_area_of_longest_diagonal_rectangle(points):
    max_diagonal_length_squared = 0
    max_area = 0

    for i in range(len(points)):
        for j in range(i + 1, len(points)):
            point1 = points[i]
            point2 = points[j]

            for k in range(len(points)):
                if k == i or k == j:
                    continue
                for l in range(k + 1, len(points)):
                    if l == i or l == j:
                        continue

                    point3 = points[k]
                    point4 = points[l]

                    # Check if the four points form a rectangle
                    if (point1[0] == point3[0] and point2[0] == point4[0] and\
                            point1[1] == point4[1] and point2[1] == point3[1]) or \
                       (point1[0] == point4[0] and point2[0] == point3[0] and\
                            point1[1] == point3[1] and point2[1] == point4[1]):

                        # Calculate diagonal length squared to avoid sqrt
                        diagonal_length_squared = (point2[0] - point1[0])**2 + (point2[1] - point1[1])**2

                        #Update the max diagonal length
                        max_diagonal_length_squared = max(max_diagonal_length_squared, diagonal_length_squared)

    for i in range(len(points)):
        for j in range(i + 1, len(points)):
            point1 = points[i]
            point2 = points[j]

            for k in range(len(points)):
                if k == i or k == j:
                    continue
                for l in range(k + 1, len(points)):
                    if l == i or l == j:
                        continue

                    point3 = points[k]
                    point4 = points[l]

                    if (point1[0] == point3[0] and point2[0] == point4[0] and\
                            point1[1] == point4[1] and point2[1] == point3[1]) or \
                       (point1[0] == point4[0] and point2[0] == point3[0] and\
                            point1[1] == point3[1] and point2[1] == point4[1]):

                        diagonal_length_squared = (point2[0] - point1[0])**2 + (point2[1] - point1[1])**2

                        #Now check if the diagonals are equal to the max
                        if diagonal_length_squared == max_diagonal_length_squared:

                            width = abs(point2[0] - point1[0])
                            height = abs(point2[1] - point1[1])
                            area = width * height

                            #Updating the max area
                            max_area = max(max_area, area)

    return max_area

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n^4) – The algorithm considers every possible pair of points (n choose 2, which is O(n^2)) as potential diagonals. For each potential diagonal, it iterates through all remaining pairs of points (again, n choose 2, or O(n^2)) to see if they can form a rectangle. This nested iteration of O(n^2) within O(n^2) leads to O(n^2 * n^2) operations. Therefore, the overall time complexity of the brute force approach is O(n^4).

Space Complexity

O(1) – The provided solution checks pairs of points and calculates rectangle properties (diagonal length, area) without storing intermediate rectangles or points in collections. It only keeps track of the maximum diagonal length and maximum area found so far, which are stored in a few variables. The number of these variables is constant and independent of the input size N (the number of points). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The goal is to find the biggest rectangle possible, considering only rectangles whose diagonal is the longest. We focus on finding the maximum area among those rectangles that share the longest possible diagonal, effectively ignoring the smaller ones to optimize our search.

Here's how the algorithm would work step-by-step:

First, find out what the longest diagonal can be based on the input rectangles by calculating the diagonal length of each rectangle using its width and height.
Keep track of the maximum diagonal length you've found so far.
Now, go through the rectangles again. This time, only consider the rectangles whose diagonal length equals the maximum diagonal length you identified.
For each of these 'longest diagonal' rectangles, calculate its area.
Keep track of the maximum area you've encountered among these special rectangles.
In the end, the maximum area you've tracked is the answer.

Code Implementation

def maximum_area_of_longest_diagonal_rectangle(rectangles):
    maximum_diagonal_length = 0

    for rectangle in rectangles:
        width, height = rectangle
        diagonal_length = (width**2 + height**2)**0.5
        maximum_diagonal_length = max(maximum_diagonal_length, diagonal_length)

    maximum_area = 0

    # Find the maximum area among rectangles with the longest diagonal.
    for rectangle in rectangles:
        width, height = rectangle
        diagonal_length = (width**2 + height**2)**0.5

        # Only consider rectangles with diagonals equal to the max diagonal.
        if diagonal_length == maximum_diagonal_length:

            current_area = width * height

            # Store the largest area from these rectangles
            maximum_area = max(maximum_area, current_area)

    return maximum_area

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of rectangles twice. The first iteration calculates the diagonal of each rectangle and finds the maximum diagonal length. This takes O(n) time. The second iteration filters for rectangles with the maximum diagonal and calculates the area, finding the maximum area among those. This second iteration also takes O(n) time. Since we perform two sequential O(n) operations, the overall time complexity is O(n) + O(n) which simplifies to O(n).

Space Complexity

O(1) – The algorithm primarily uses a few variables to store the maximum diagonal length and the maximum area encountered so far. The steps involve iterating through the rectangles, calculating diagonals and areas, and updating the maximums. No auxiliary data structures that scale with the number of rectangles (N) are employed, therefore the space used remains constant regardless of the input size.

Edge Cases

Case	How to Handle
Null or empty dimensions array	Return 0 immediately since no rectangles are present.
Dimensions array with a single rectangle	Calculate and return the area of the single rectangle.
Rectangles with zero length or width	The area will be zero; the diagonal calculation must handle zero values correctly, and the max area should still be handled correctly.
Rectangles with very large lengths or widths causing potential integer overflow when calculating area or diagonal (diagonal using multiplication)	Use long data type for intermediate calculations of area and diagonal to avoid overflow.
Multiple rectangles with the same longest diagonal	Keep track of the maximum area among the rectangles with the longest diagonal.
Dimensions with negative values	Return 0 or throw an IllegalArgumentException since negative lengths and widths are invalid.
Very large number of rectangles to ensure algorithm efficiency, and potentially exhausting memory	The algorithm should iterate through the array once, so the time complexity is O(N), and memory usage is constant.
Rectangles with extremely close diagonals; compare doubles with a tolerance	When comparing diagonals, use a small epsilon for floating-point comparisons (e.g., Math.abs(d1 - d2) < 1e-9).