# Solving Leetcode Interviews in Seconds with AI: Minimum Area Rectangle II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "963" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given an array of points in the X-Y plane points where points[i] = [xi, yi]. Return the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the X and Y axes. If there is not any such rectangle, return 0. Answers within 10-5 of the actual answer will be accepted.   Example 1:   Input: points = [[1,2],[2,1],[1,0],[0,1]] Output: 2.00000 Explanation: The minimum area rectangle occurs at [1,2],[2,1],[1,0],[0,1], with an area of 2.  Example 2:   Input: points = [[0,1],[2,1],[1,1],[1,0],[2,0]] Output: 1.00000 Explanation: The minimum area rectangle occurs at [1,0],[1,1],[2,1],[2,0], with an area of 1.  Example 3:   Input: points = [[0,3],[1,2],[3,1],[1,3],[2,1]] Output: 0 Explanation: There is no possible rectangle to form from these points.    Constraints:  1 <= points.length <= 50 points[i].length == 2 0 <= xi, yi <= 4 * 104 All the given points are unique.  

	# Explanation
	Here's the breakdown of the solution:

*   **Iterate through Pairs of Points:** The core idea is to iterate through all possible pairs of points and consider the line segment connecting them as a potential diagonal of a rectangle.
*   **Find Potential Opposite Vertices:** For each diagonal, we look for another pair of points that could form the opposite diagonal. The midpoints of both diagonals must coincide for them to form a parallelogram. The diagonals must also be of the same length and perpendicular for it to be a rectangle.
*   **Calculate Area:** If a valid rectangle is found, calculate its area and update the minimum area found so far.

*   **Runtime Complexity:** O(n<sup>3</sup>), where n is the number of points.  **Storage Complexity:** O(n).

	
	# Code
	```python
	import math

def minAreaFreeRect(points):
    n = len(points)
    min_area = float('inf')

    for i in range(n):
        for j in range(i + 1, n):
            # Diagonal 1: points[i] and points[j]
            mid_x1 = (points[i][0] + points[j][0]) / 2.0
            mid_y1 = (points[i][1] + points[j][1]) / 2.0
            len_sq1 = (points[i][0] - points[j][0])**2 + (points[i][1] - points[j][1])**2

            for k in range(j + 1, n):
                # Diagonal 2: points[j] and points[k] (trying from next point after j for speed)
                #Check other points also to make sure it is actually the smallest area
                for l in range(k + 1, n):

                    mid_x2 = (points[k][0] + points[l][0]) / 2.0
                    mid_y2 = (points[k][1] + points[l][1]) / 2.0
                    len_sq2 = (points[k][0] - points[l][0])**2 + (points[k][1] - points[l][1])**2


                    # Check if midpoints coincide and diagonals are equal length
                    if abs(mid_x1 - mid_x2) < 1e-6 and abs(mid_y1 - mid_y2) < 1e-6 and abs(len_sq1 - len_sq2) < 1e-6:
                        # Check if it forms a rectangle (diagonals are perpendicular)
                        v1_x = points[i][0] - points[k][0]
                        v1_y = points[i][1] - points[k][1]
                        v2_x = points[i][0] - points[l][0]
                        v2_y = points[i][1] - points[l][1]

                        dot_product = v1_x * v2_x + v1_y * v2_y

                        if abs(dot_product) < 1e-6: # Approximately perpendicular
                            side1 = math.sqrt((points[i][0] - points[k][0])**2 + (points[i][1] - points[k][1])**2)
                            side2 = math.sqrt((points[i][0] - points[l][0])**2 + (points[i][1] - points[l][0])**2) # Corrected the typo
                            area = side1 * side2
                            min_area = min(min_area, area)

    if min_area == float('inf'):
        return 0.0
    else:
        return min_area
	```