LeetCode 3380: Maximum Area Rectangle With Point Constraints I Solution

Master LeetCode problem 3380 (Maximum Area Rectangle With Point Constraints I), 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.

3380. Maximum Area Rectangle With Point Constraints I

Medium

Array Math Binary Indexed Tree Segment Tree Geometry Sorting Enumeration

Problem Explanation

Explanation:

To solve this problem, we can iterate through all possible pairs of points and calculate the Manhattan distance between them. If the Manhattan distance is equal to the diagonal of the rectangle formed by those two points, we can calculate the area of the rectangle and keep track of the maximum area found so far. We need to ensure that no other point lies inside or on the border of the rectangle.

Algorithm:
1. Iterate through all pairs of points.
2. For each pair of points, calculate the Manhattan distance.
3. If the Manhattan distance is equal to the diagonal of the rectangle, check if there is no other point inside or on the border.
4. Calculate the area of the rectangle and update the maximum area found so far.
5. Return the maximum area or -1 if no valid rectangle is found.
Time Complexity: O(n^2) where n is the number of points.
Space Complexity: O(1)

Solution Code

class Solution {
    public int maxArea(int[][] points) {
        int maxArea = -1;
        for (int i = 0; i < points.length; i++) {
            for (int j = i + 1; j < points.length; j++) {
                int manhattanDist = Math.abs(points[i][0] - points[j][0]) + Math.abs(points[i][1] - points[j][1]);
                int area = manhattanDist * manhattanDist;
                boolean validRectangle = true;
                for (int[] point : points) {
                    if (point != points[i] && point != points[j] &&
                            point[0] >= Math.min(points[i][0], points[j][0]) &&
                            point[0] <= Math.max(points[i][0], points[j][0]) &&
                            point[1] >= Math.min(points[i][1], points[j][1]) &&
                            point[1] <= Math.max(points[i][1], points[j][1])) {
                        validRectangle = false;
                        break;
                    }
                }
                if (validRectangle) {
                    maxArea = Math.max(maxArea, area);
                }
            }
        }
        return maxArea;
    }
}

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