LeetCode-in-Java

3047. Find the Largest Area of Square Inside Two Rectangles

Medium

There exist n rectangles in a 2D plane. You are given two 0-indexed 2D integer arrays bottomLeft and topRight, both of size n x 2, where bottomLeft[i] and topRight[i] represent the bottom-left and top-right coordinates of the ith rectangle respectively.

You can select a region formed from the intersection of two of the given rectangles. You need to find the largest area of a square that can fit inside this region if you select the region optimally.

Return the largest possible area of a square, or 0 if there do not exist any intersecting regions between the rectangles.

Example 1:

Input: bottomLeft = [[1,1],[2,2],[3,1]], topRight = [[3,3],[4,4],[6,6]]

Output: 1

Explanation: A square with side length 1 can fit inside either the intersecting region of rectangle 0 and rectangle 1, or the intersecting region of rectangle 1 and rectangle 2. Hence the largest area is side * side which is 1 * 1 == 1.

It can be shown that a square with a greater side length can not fit inside any intersecting region.

Example 2:

Input: bottomLeft = [[1,1],[2,2],[1,2]], topRight = [[3,3],[4,4],[3,4]]

Output: 1

Explanation: A square with side length 1 can fit inside either the intersecting region of rectangle 0 and rectangle 1, the intersecting region of rectangle 1 and rectangle 2, or the intersection region of all 3 rectangles. Hence the largest area is side * side which is 1 * 1 == 1.

It can be shown that a square with a greater side length can not fit inside any intersecting region. Note that the region can be formed by the intersection of more than 2 rectangles.

Example 3:

Input: bottomLeft = [[1,1],[3,3],[3,1]], topRight = [[2,2],[4,4],[4,2]]

Output: 0

Explanation: No pair of rectangles intersect, hence, we return 0.

Constraints:

• n == bottomLeft.length == topRight.length
• 2 <= n <= 103
• bottomLeft[i].length == topRight[i].length == 2
• 1 <= bottomLeft[i][0], bottomLeft[i][1] <= 107
• 1 <= topRight[i][0], topRight[i][1] <= 107
• bottomLeft[i][0] < topRight[i][0]
• bottomLeft[i][1] < topRight[i][1]

Solution

public class Solution {
    public long largestSquareArea(int[][] bottomLeft, int[][] topRight) {
        int n = bottomLeft.length;
        long maxArea = 0;
        for (int i = 0; i < n; i++) {
            int ax = bottomLeft[i][0];
            int ay = bottomLeft[i][1];
            int bx = topRight[i][0];
            int by = topRight[i][1];
            for (int j = i + 1; j < n; j++) {
                int cx = bottomLeft[j][0];
                int cy = bottomLeft[j][1];
                int dx = topRight[j][0];
                int dy = topRight[j][1];
                int x1 = Math.max(ax, cx);
                int y1 = Math.max(ay, cy);
                int x2 = Math.min(bx, dx);
                int y2 = Math.min(by, dy);
                int minSide = Math.min(x2 - x1, y2 - y1);
                long area = (long) Math.pow(Math.max(minSide, 0), 2);
                maxArea = Math.max(maxArea, area);
            }
        }
        return maxArea;
    }
}

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