Minimum Area Rectangle Ii - Leetcode Solution

Problem Statement:

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.

Solution:

Approach 1: Brute Force

One way of solving this problem is to consider all possible pairs of points and check if they form a rectangle with sides parallel to the x and y axes. We can then calculate the area of each possible rectangle and return the minimum area we find.

However, the time complexity of this approach would be O(n^4), which is too high and not feasible for large inputs.

Approach 2: Hashing

We can optimize the brute force approach by using hashing. We can store all the points in a hash set and for each pair of points, we can check if their corresponding opposite vertices are also in the set.

Let's say we have two points A and B, where A = (x1, y1) and B = (x2, y2). The other two vertices of the rectangle would be C and D, where C = (x2, y1) and D = (x1, y2).

We can check if C and D are in the set using the following code:

if (set.contains(new Point(x2, y1)) && set.contains(new Point(x1, y2))) {...}

We can then calculate the area of the rectangle formed by A, B, C, and D and return the minimum area we find.

The time complexity of this approach would be O(n^2), which is much better than the brute force approach.

Approach 3: Rotating Calipers

The rotating calipers algorithm can be used to solve this problem efficiently in O(n log n) time.

The algorithm works by first finding the convex hull of the set of points. Then, for each pair of points on the convex hull, we can calculate the distance between them and rotate the hull so that the line connecting these two points is parallel to the x-axis.

We can then find the smallest rectangle that can be formed with sides parallel to the x and y axes and passing through these two points. We can repeat this process for all pairs of points on the convex hull and return the minimum area of all the rectangles we find.

The time complexity of this approach is O(n log n) because finding the convex hull takes O(n log n) time and then we need to consider all pairs of points on the convex hull, which takes O(n) time.

Conclusion:

We can solve the Minimum Area Rectangle II problem using different approaches. The brute force approach has a time complexity of O(n^4), which is not feasible for large inputs. The hashing approach optimizes the brute force approach by using a hash set to check if the opposite vertices are in the set, and has a time complexity of O(n^2). The rotating calipers algorithm provides the most efficient solution with a time complexity of O(n log n).

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