Solution Explanation: Minimum Area Rectangle II

This problem asks to find the minimum area of a rectangle formed by any four points from a given set of points, where the sides of the rectangle are not necessarily parallel to the x and y axes.

Approach:

The solution uses a brute-force approach with optimizations. It iterates through all possible combinations of four points. For each combination, it checks if they form a rectangle and calculates its area. The minimum area found is returned. Optimizations include:

1. Preprocessing: A set s is created to store the points for efficient lookups (checking if a potential fourth point exists). This reduces the time complexity compared to searching through the list of points repeatedly.

2. Orthogonality Check: Instead of directly checking if four points form a rectangle, we check for orthogonality of vectors. If vectors formed by three points (v21 and v31) are orthogonal (their dot product is zero), then it implies the existence of a rectangle.

3. Area Calculation: Once orthogonality is confirmed, the area is calculated using the magnitudes of the orthogonal vectors (width and height of the rectangle).

Code Breakdown (Python3 Example):

class Solution:
    def minAreaFreeRect(self, points: List[List[int]]) -> float:
        s = {(x, y) for x, y in points}  # Set for efficient point lookup
        n = len(points)
        ans = float('inf')  # Initialize minimum area to infinity
 
        for i in range(n):
            x1, y1 = points[i]
            for j in range(n): # Iterate through all pairs of points
                if j != i:
                    x2, y2 = points[j]
                    for k in range(j + 1, n): # Iterate through the remaining points
                        if k != i:
                            x3, y3 = points[k]
                            x4 = x2 - x1 + x3  # Calculate potential fourth point coordinates
                            y4 = y2 - y1 + y3
                            if (x4, y4) in s:  # Check if fourth point exists
                                v21 = (x2 - x1, y2 - y1)  # Vector from point 1 to point 2
                                v31 = (x3 - x1, y3 - y1)  # Vector from point 1 to point 3
                                if v21[0] * v31[0] + v21[1] * v31[1] == 0: # Check orthogonality (dot product = 0)
                                    w = math.sqrt(v21[0] ** 2 + v21[1] ** 2)  # Calculate width
                                    h = math.sqrt(v31[0] ** 2 + v31[1] ** 2)  # Calculate height
                                    ans = min(ans, w * h)  # Update minimum area
        return 0 if ans == float('inf') else ans # Return 0 if no rectangle found

Time Complexity Analysis:

The nested loops iterate through all combinations of three points, which leads to a time complexity of O(n³), where n is the number of points. The set lookup (in s) takes O(1) on average. Other operations inside the loops have constant time complexity.

Space Complexity Analysis:

The space complexity is dominated by the set s, which stores n points. Therefore, the space complexity is O(n).

The provided code in other languages (Java, C++, Go, TypeScript) follows the same algorithmic approach, differing only in syntax and data structure implementations. The time and space complexities remain the same.