LeetCode 1459: Rectangles Area Solution

Master LeetCode problem 1459 (Rectangles Area), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1459. Rectangles Area

Medium

Database

Problem Explanation

Explanation:

To solve this problem, we need to find the total area covered by a set of rectangles. We can do this by first creating a list of all the x-coordinates where the rectangles start or end, sorted in ascending order. Then, we iterate through these x-coordinates and calculate the total area covered by rectangles at each segment.

Create a list of x-coordinates:
- Add the x-coordinates of the left and right edges of each rectangle to a list.
- Sort the list in ascending order to get a sorted list of x-coordinates.
Iterate through the sorted x-coordinates:
- At each segment between two x-coordinates, calculate the total height covered by rectangles.
- Multiply the segment width by the total height to get the area covered by rectangles in that segment.
- Update the total area by adding the area of the current segment.
Return the total area as the final result.

Time Complexity: O(nlogn)
Space Complexity: O(n)

: :

Solution Code

import java.util.*;

class Solution {
    public int rectanglesArea(int[][] rectangles) {
        List<Integer> xCoordinates = new ArrayList<>();
        for (int[] rect : rectangles) {
            xCoordinates.add(rect[0]);
            xCoordinates.add(rect[2]);
        }
        Collections.sort(xCoordinates);

        long totalArea = 0;
        for (int i = 0; i < xCoordinates.size() - 1; i++) {
            int start = xCoordinates.get(i);
            int end = xCoordinates.get(i + 1);
            int height = 0;
            for (int[] rect : rectangles) {
                if (rect[0] <= start && rect[2] >= end) {
                    height = Math.max(height, rect[1]);
                }
            }
            totalArea += (long) (end - start) * height;
        }

        return (int) (totalArea % 1000000007);
    }
}

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