Leetcode #963: Minimum Area Rectangle II
In this guide, we solve Leetcode #963 Minimum Area Rectangle II in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Geometry, Array, Hash Table, Math
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
class Solution:
def minAreaFreeRect(self, points: List[List[int]]) -> float:
s = {(x, y) for x, y in points}
n = len(points)
ans = inf
for i in range(n):
x1, y1 = points[i]
for j in range(n):
if j != i:
x2, y2 = points[j]
for k in range(j + 1, n):
if k != i:
x3, y3 = points[k]
x4 = x2 - x1 + x3
y4 = y2 - y1 + y3
if (x4, y4) in s:
v21 = (x2 - x1, y2 - y1)
v31 = (x3 - x1, y3 - y1)
if v21[0] * v31[0] + v21[1] * v31[1] == 0:
w = sqrt(v21[0] ** 2 + v21[1] ** 2)
h = sqrt(v31[0] ** 2 + v31[1] ** 2)
ans = min(ans, w * h)
return 0 if ans == inf else ans
Complexity
The time complexity is and the space complexity is , where is the length of the array . The space complexity is , where is the length of the array .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.