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


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "939" 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 a rectangle formed from these points, with sides parallel to the X and Y axes. If there is not any such rectangle, return 0.   Example 1:   Input: points = [[1,1],[1,3],[3,1],[3,3],[2,2]] Output: 4  Example 2:   Input: points = [[1,1],[1,3],[3,1],[3,3],[4,1],[4,3]] Output: 2    Constraints:  1 <= points.length <= 500 points[i].length == 2 0 <= xi, yi <= 4 * 104 All the given points are unique.  

	# Explanation
	Here's the breakdown of the approach, complexities, and the Python code for finding the minimum area rectangle:

*   **High-Level Approach:**
    *   Iterate through all possible pairs of points to consider them as potential corners of a rectangle.
    *   For each pair, check if the other two corners required to form a rectangle (with sides parallel to the axes) exist in the given set of points.
    *   If all four corners exist, calculate the area of the rectangle and update the minimum area found so far.

*   **Complexity:**
    *   Runtime Complexity: O(n^2), where n is the number of points.
    *   Storage Complexity: O(n), mainly for the set used to efficiently check the existence of points.

	
	# Code
	```python
	def minAreaRect(points):
    """
    Finds the minimum area of a rectangle formed from the given points,
    with sides parallel to the X and Y axes.

    Args:
      points: A list of points in the X-Y plane.

    Returns:
      The minimum area of a rectangle, or 0 if no such rectangle exists.
    """

    n = len(points)
    if n < 4:
        return 0

    point_set = set((x, y) for x, y in points)
    min_area = float('inf')

    for i in range(n):
        for j in range(i + 1, n):
            x1, y1 = points[i]
            x2, y2 = points[j]

            # Check if the points form a diagonal (not a side)
            if x1 != x2 and y1 != y2:
                # Check if the other two corners exist
                if (x1, y2) in point_set and (x2, y1) in point_set:
                    area = abs(x1 - x2) * abs(y1 - y2)
                    min_area = min(min_area, area)

    if min_area == float('inf'):
        return 0
    else:
        return min_area
	```