Rectangles Area - Leetcode Solution

The Rectangles Area problem on LeetCode can be solved using a brute-force approach to check all possible pairs of rectangles and calculate their intersection area. The problem statement is as follows:

You are given an array rectangles where rectangles[i] = [ai, bi] represents the ith rectangle of length ai and width bi. You are also given an integer k which represents the number of rectangles that can be placed end to end horizontally without overlapping. Return the maximum area of a valid placement of rectangles.

To solve the problem in an optimized manner, follow these steps:

1. Sort the given rectangles array in descending order of their widths.

2. Initialize two variables i and j, where i stores the number of rectangles placed so far, and j is used as a pointer to traverse through the sorted rectangles in descending order of their widths.

3. Initialize two variables, width and height, which represents the width and height of the current placement of rectangles.

4. Loop through the sorted array rectangles, starting from index j=0 to n-1 (where n is the length of rectangles array)

   a. Check if the current placement of rectangles is valid, which means that the number of rectangles placed so far (i) is less than or equal to k.

   b. Calculate the updated value of width and height, where width is the sum of the widths of the placed rectangles, and height is the maximum height of the placed rectangles.

   c. If the placement is valid, update the maximum area so far using the formula, maxArea = max(maxArea, height * width).

   d. If the placement is not valid, break out of the loop.

   e. Increment the value of j and i accordingly.

5. Return the maximum area calculated in step 4.

Here is the Python code to implement the above algorithm:

class Solution:
    def maxSumSubmatrix(self, rectangles: List[List[int]], k: int) -> int:
        n = len(rectangles)
        rectangles.sort(key=lambda x: x[1], reverse=True)

        maxArea = float('-inf')
        for j in range(n):
            width, height, i = 0, 0, 0
            
            while i + j < n and (i+1)*width <= k:
                width += rectangles[i+j][0]
                height = max(height, rectangles[i+j][1])
                
                if (i+1)*width <= k:
                    maxArea = max(maxArea, height*width)
                    
                i += 1
        
        return maxArea

The time complexity of this solution is O(n^2 log n), where n is the length of the rectangles array, due to the sorting operation performed on the array. And the space complexity is O(1) as we are not using any additional data structures.

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