What is the time complexity of Minimum Area Rectangle II?

The optimal solution has a time complexity of O(N^4).

What is the space complexity of Minimum Area Rectangle II?

The space complexity is O(N^2).

What data structures are used to solve Minimum Area Rectangle II?

This problem uses Array, Hash Table, Math, Geometry. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Area Rectangle II?

Insight 1: A rectangle is defined by two diagonals that have the same midpoint and the same length. Insight 2: We can use a hash table (or dictionary) to store the center and squared length of diagonals. This allows efficient lookup of diagonals with the same properties.

Which companies ask LeetCode 963. Minimum Area Rectangle II in interviews?

This problem is frequently asked at Verily.

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Minimum Area Rectangle II - LeetCode 963 Solution

Minimum Area Rectangle II - Complete Solution Guide

Minimum Area Rectangle II is LeetCode problem 963, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an array of points in the X-Y plane points where points[i] = [x i , y i ] . 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: Inpu

Detailed Explanation

The problem asks us to find the minimum area of any rectangle that can be formed from a given set of points in the 2D plane. The rectangle's sides do not necessarily have to be parallel to the X and Y axes, meaning we need to consider rotated rectangles. If no rectangle can be formed, we should return 0.0. The solution needs to be accurate to within 10^-5.

Solution Approach

The solution iterates through all possible pairs of points, treating each pair as a potential diagonal of a rectangle. It calculates the midpoint and squared length of this diagonal. These two values (midpoint and squared length) are used as a key in a hash table. If another diagonal with the same key is found, it means we have two diagonals with the same midpoint and length. These two diagonals, if they don't share a common point, potentially form a rectangle. We then check if these two diagonals can be used to calculate a minimum area of a valid rectangle and update accordingly.

Step-by-Step Algorithm

Step 1: Iterate through all possible pairs of points (i, j) where i < j. Each pair represents a potential diagonal.
Step 2: Calculate the midpoint of the diagonal formed by points[i] and points[j]. The midpoint coordinates are (x1 + x2)/2 and (y1 + y2)/2. Since the values can be decimals, keep center_x_2 and center_y_2 as the sum.
Step 3: Calculate the squared length of the diagonal (d_sq) using the distance formula: (x1 - x2)^2 + (y1 - y2)^2.
Step 4: Store the diagonal (points[i], points[j]) in a hash table. The key for the hash table is a combination of the midpoint coordinates (center_x_2, center_y_2) and the squared length (d_sq).
Step 5: Iterate through all the diagonals stored for each key in the hash table. For each key, compare all the pairs of diagonals. Only consider entries with 2 or more diagonals
Step 6: For each pair of diagonals (p1, p2) and (p3, p4), calculate the squared lengths of the sides of the potential rectangle. This is done by calculating the squared distance between p1 and p3 (dist1_sq) and the squared distance between p1 and p4 (dist2_sq).
Step 7: Calculate the squared area (area_sq) by multiplying dist1_sq and dist2_sq.
Step 8: Update the minimum squared area (min_area_sq) if area_sq is smaller than the current minimum and area_sq is greater than 0 (to avoid considering degenerate cases).
Step 9: After iterating through all possible pairs of diagonals, if min_area_sq is still infinity, return 0.0 (no rectangle found).
Step 10: Otherwise, return the square root of min_area_sq to get the actual minimum area.

Key Insights

Insight 1: A rectangle is defined by two diagonals that have the same midpoint and the same length.
Insight 2: We can use a hash table (or dictionary) to store the center and squared length of diagonals. This allows efficient lookup of diagonals with the same properties.
Insight 3: Instead of calculating the actual area directly, we calculate the square of the area to avoid floating-point precision issues until the end.

Complexity Analysis

Time Complexity: O(N^4)

Space Complexity: O(N^2)

Topics

This problem involves: Array, Hash Table, Math, Geometry.

Companies

Asked at: Verily.

Minimum Area Rectangle II | LeetCode 963

Medium

Array Hash Table Math Geometry

Time: O(N^4)Space: O(N^2)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public double minAreaFreeRect(int[][] points) {
        int n = points.length;
        if (n < 4) {
            return 0.0;
        }

        Map<String, List<int[][]>> diagonals = new HashMap<>();
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int[] p1 = points[i];
                int[] p2 = points[j];

                int center_x_2 = p1[0] + p2[0];
                int center_y_2 = p1[1] + p2[1];
                int d_sq = (p1[0] - p2[0]) * (p1[0] - p2[0]) + (p1[1] - p2[1]) * (p1[1] - p2[1]);

                String key = center_x_2 + "," + center_y_2 + "," + d_sq;
                if (!diagonals.containsKey(key)) {
                    diagonals.put(key, new ArrayList<>());
                }
                int[][] pair = {p1, p2};
                diagonals.get(key).add(pair);
            }
        }

        double min_area_sq = Double.POSITIVE_INFINITY;

        for (List<int[][]> pairs : diagonals.values()) {
            int m = pairs.size();
            if (m < 2) {
                continue;
            }

            for (int i = 0; i < m; i++) {
                for (int j = i + 1; j < m; j++) {
                    int[] p1 = pairs.get(i)[0];
                    int[] p2 = pairs.get(i)[1];
                    int[] p3 = pairs.get(j)[0];
                    int[] p4 = pairs.get(j)[1];

                    int dist1_sq = (p1[0] - p3[0]) * (p1[0] - p3[0]) + (p1[1] - p3[1]) * (p1[1] - p3[1]);
                    int dist2_sq = (p1[0] - p4[0]) * (p1[0] - p4[0]) + (p1[1] - p4[1]) * (p1[1] - p4[1]);

                    double area_sq = (double) dist1_sq * (double) dist2_sq;
                    if (area_sq > 0) {
                        min_area_sq = Math.min(min_area_sq, area_sq);
                    }
                }
            }
        }

        if (min_area_sq == Double.POSITIVE_INFINITY) {
            return 0.0;
        } else {
            return Math.sqrt(min_area_sq);
        }
    }
}

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Problem Info

Description

Problem 963: Minimum Area Rectangle II - Detailed explanation and solution approaches.

Topics

Array Hash Table Math Geometry

Companies

Verily

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:963

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Iterate through all possible pairs of points (i, j) where i < j. Each pair represents a potential diagonal.
2
Step 2: Calculate the midpoint of the diagonal formed by points[i] and points[j]. The midpoint coordinates are (x1 + x2)/2 and (y1 + y2)/2. Since the values can be decimals, keep center_x_2 and center_y_2 as the sum.
3
Step 3: Calculate the squared length of the diagonal (d_sq) using the distance formula: (x1 - x2)^2 + (y1 - y2)^2.
4
Step 4: Store the diagonal (points[i], points[j]) in a hash table. The key for the hash table is a combination of the midpoint coordinates (center_x_2, center_y_2) and the squared length (d_sq).
5
Step 5: Iterate through all the diagonals stored for each key in the hash table. For each key, compare all the pairs of diagonals. Only consider entries with 2 or more diagonals
6
Step 6: For each pair of diagonals (p1, p2) and (p3, p4), calculate the squared lengths of the sides of the potential rectangle. This is done by calculating the squared distance between p1 and p3 (dist1_sq) and the squared distance between p1 and p4 (dist2_sq).
7
Step 7: Calculate the squared area (area_sq) by multiplying dist1_sq and dist2_sq.
8
Step 8: Update the minimum squared area (min_area_sq) if area_sq is smaller than the current minimum and area_sq is greater than 0 (to avoid considering degenerate cases).
9
Step 9: After iterating through all possible pairs of diagonals, if min_area_sq is still infinity, return 0.0 (no rectangle found).
10
Step 10: Otherwise, return the square root of min_area_sq to get the actual minimum area.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(N^4)

💾

Space Complexity

O(N^2)

📝

Why?

The nested loops to find all pairs of points contribute O(N^2). The hash table stores O(N^2) data. The innermost loops which check for valid rectangles take another O(N^2), resulting in overall O(N^4) time complexity and O(N^2) space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force with Angle Checking

Iterate through all combinations of 4 points and check if they form a rectangle. For each combination, check if the angles are right angles and if the opposite sides are equal in length.

Time: O(N^4)Space: O(1)

Trade-offs: This approach has the same time complexity as the main solution, but is likely to be slower in practice due to the more complex calculations required to check for right angles. It might be preferred if memory usage is a strict constraint.

✨

Key Insights

1Insight 1: A rectangle is defined by two diagonals that have the same midpoint and the same length.
2Insight 2: We can use a hash table (or dictionary) to store the center and squared length of diagonals. This allows efficient lookup of diagonals with the same properties.
3Insight 3: Instead of calculating the actual area directly, we calculate the square of the area to avoid floating-point precision issues until the end.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: Less than 4 points. The solution handles this by returning 0.0 directly.
Edge case 2: No rectangle can be formed. The solution handles this by initializing the minimum area to infinity and returning 0.0 if it remains infinity after all calculations.
Edge case 3: Collinear points. The solution handles this correctly as it will not find two diagonals with the same center and length to form a rectangle.
Edge case 4: Degenerate rectangles (area = 0). The solution ensures the area is greater than zero before updating the minimum area.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. When calculating squared distances, the result can easily overflow the integer range. Use `long` or `double` to prevent this.

✗Common mistake 2: Floating-point precision errors. Directly comparing floating-point numbers for equality can be unreliable. Use a tolerance value (e.g., 10^-5) to check if the numbers are approximately equal or work with squared values to avoid the square root operation early on.

✗Common mistake 3: Forgetting to handle the case where no rectangle is found (returning 0.0).

✗Common mistake 4: Not handling memory deallocation in C. Memory allocated for hashmap and all elements should be freed to avoid memory leaks.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Image processing - Identifying rectangular objects in images (e.g., screens, documents).

🏦Real-world scenario 2: Computer vision - Object detection and recognition in video streams.

🎮Real-world scenario 3: Geographic Information Systems (GIS) - Finding rectangular plots of land or buildings on a map.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 3, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Area Rectangle II"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?