939. Minimum Area Rectangle

Given a set of points in the xy-plane, determine the minimum area of a rectangle formed from these points, with sides parallel to the x and y axes.

If there isn't any rectangle, return 0.

 

Example 1:

Input: [[1,1],[1,3],[3,1],[3,3],[2,2]]
Output: 4 

Example 2:

Input: [[1,1],[1,3],[3,1],[3,3],[4,1],[4,3]]
Output: 2 

 

 

Approach 2: Count by Diagonal

Intuition

For each pair of points in the array, consider them to be the long diagonal of a potential rectangle. We can check if all 4 points are there using a Set.

For example, if the points are (1, 1) and (5, 5), we check if we also have (1, 5) and (5, 1). If we do, we have a candidate rectangle.

Algorithm

Put all the points in a set. For each pair of points, if the associated rectangle are 4 distinct points all in the set, then take the area of this rectangle as a candidate answer.

 

class Solution {
    public int minAreaRect(int[][] points) {
        Set<Integer> pointSet = new HashSet();
        for (int[] point: points)
            pointSet.add(40001 * point[0] + point[1]);

        int ans = Integer.MAX_VALUE;
        for (int i = 0; i < points.length; ++i)
            for (int j = i+1; j < points.length; ++j) {
                if (points[i][0] != points[j][0] && points[i][1] != points[j][1]) {
                    if (pointSet.contains(40001 * points[i][0] + points[j][1]) &&
                            pointSet.contains(40001 * points[j][0] + points[i][1])) {
                        ans = Math.min(ans, Math.abs(points[j][0] - points[i][0]) *
                                            Math.abs(points[j][1] - points[i][1]));
                    }
                }
            }

        return ans < Integer.MAX_VALUE ? ans : 0;
    }
}

Complexity Analysis

• Time Complexity: O(N^2), where NN is the length of points.

• Space Complexity: O(N), where HH is the height of the tree.