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Given n rectangles represented as a list of four integers [x1, y1, x2, y2] where (x1, y1) is the bottom-left corner and (x2, y2) is the top-right corner of the rectangle, determine the maximum area of the rectangle that can be formed from the longest diagonal of any rectangles.

Clarifying Questions

  1. Diagonal Calculation: Do we need to consider only the originally given rectangles, or any potential rectangles formed by the coordinates?
  2. Overlap / Combination: Should we consider the union or intersection of rectangles for forming diagonals or areas?
  3. Input Validation: What should be assumed about the input? For example, can we assume all coordinates are integers and x1 < x2 and y1 < y2 for every rectangle?
  4. Bounding Box: Are there any maximum constraints on the coordinate values?

Strategy

The problem essentially boils down to two main steps:

  1. Determine the Longest Diagonal: Calculate the length of the diagonals for each rectangle. The formula for the diagonal length (d) considering two points (x1, y1) and (x2, y2) is d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
  2. Calculate the Area: Once the rectangle with the longest diagonal is determined, calculate its area using the formula area = (x2 - x1) * (y2 - y1).

Code

Here’s a Python function to solve the problem:

import math

def max_area_longest_diagonal(rectangles):
    max_diagonal = 0
    max_area = 0
    
    for rect in rectangles:
        x1, y1, x2, y2 = rect
        diagonal_length = math.sqrt((x2 - x1) ** 2 + (y2 - y1) ** 2)
        area = (x2 - x1) * (y2 - y1)
        
        if diagonal_length > max_diagonal:
            max_diagonal = diagonal_length
            max_area = area
    
    return max_area

# Example usage:
rectangles = [
    [1, 1, 4, 5],
    [2, 3, 8, 7],
    [3, 2, 5, 6]
]

print(max_area_longest_diagonal(rectangles)) # Output would be the area of rectangle with the longest diagonal

Time Complexity

The time complexity for this solution is:

Each individual operation within the loop (calculating the diagonal, area, and checking conditions) is constant time, O(1). Therefore, the overall time complexity remains O(n).

This should ensure efficient handling even when n is large.

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