What is the time complexity of Minimum Area Rectangle?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Minimum Area Rectangle?

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

What are the key insights for solving Minimum Area Rectangle?

Insight 1: To form a rectangle with sides parallel to the axes, we need to find four points such that two pairs of points have the same x-coordinate and the other two pairs have the same y-coordinate. Insight 2: Using a set or hash table to store the points allows for efficient checking of whether a potential rectangle's corners exist within the input point set.

Which companies ask LeetCode 939. Minimum Area Rectangle in interviews?

This problem is frequently asked at ByteDance, Flipkart, Snap, Verily.

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Minimum Area Rectangle - LeetCode 939 Solution

Q: How to solve Minimum Area Rectangle?

The solution iterates through all possible pairs of points in the input array. For each pair, it checks if these points can form two corners of a rectangle with sides parallel to the axes (i.e., they do not have the same x or y coordinates). If they can, it checks if the other two corners of the rectangle exist within the given set of points. If all four corners exist, the area of the rectangle is calculated, and the minimum area is updated accordingly. A set is used to efficiently check if a point exists.

Minimum Area Rectangle - Complete Solution Guide

Minimum Area Rectangle is LeetCode problem 939, 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 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 <= x i , y i <= 4 * 10 4 All the given points are

Detailed Explanation

The problem asks us to find the minimum area of a rectangle formed by a given set of points in the XY plane, with the constraint that the rectangle's sides must be parallel to the X and Y axes. If no such rectangle can be formed from the given points, we should return 0. The input is an array of points, where each point is represented by a pair of coordinates (x, y).

Solution Approach

The solution iterates through all possible pairs of points in the input array. For each pair, it checks if these points can form two corners of a rectangle with sides parallel to the axes (i.e., they do not have the same x or y coordinates). If they can, it checks if the other two corners of the rectangle exist within the given set of points. If all four corners exist, the area of the rectangle is calculated, and the minimum area is updated accordingly. A set is used to efficiently check if a point exists.

Step-by-Step Algorithm

Step 1: Create a set (or hash table) containing all the given points for efficient lookup.
Step 2: Iterate through all possible pairs of points (p1, p2) from the input array.
Step 3: For each pair of points (p1(x1, y1), p2(x2, y2)), check if x1 != x2 and y1 != y2. This condition ensures that the two points can be diagonally opposite corners of a rectangle.
Step 4: If the condition in Step 3 is met, check if the other two potential corners of the rectangle, (x1, y2) and (x2, y1), are present in the point set.
Step 5: If both corners (x1, y2) and (x2, y1) exist in the set, calculate the area of the rectangle as abs(x1 - x2) * abs(y1 - y2).
Step 6: Update the minimum area found so far if the calculated area is smaller.
Step 7: After iterating through all pairs of points, return the minimum area. If no rectangle was found, return 0.

Key Insights

Insight 1: To form a rectangle with sides parallel to the axes, we need to find four points such that two pairs of points have the same x-coordinate and the other two pairs have the same y-coordinate.
Insight 2: Using a set or hash table to store the points allows for efficient checking of whether a potential rectangle's corners exist within the input point set.
Insight 3: We can iterate through all possible pairs of points and check if they can be two corners of a rectangle. If they are, we check if the other two corners exist in the set of points.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

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

Companies

Asked at: ByteDance, Flipkart, Snap, Verily.

Minimum Area Rectangle | LeetCode 939

Medium

Array Hash Table Math Geometry Sorting

Time: O(n^2)Space: O(n)

Solution Code

Available in:

import java.util.HashSet;
import java.util.Set;

class Solution {
    public int minAreaRect(int[][] points) {
        Set<String> pointSet = new HashSet<>();
        for (int[] point : points) {
            pointSet.add(point[0] + "," + point[1]);
        }

        int minArea = Integer.MAX_VALUE;

        for (int i = 0; i < points.length; i++) {
            for (int j = i + 1; j < points.length; j++) {
                int x1 = points[i][0];
                int y1 = points[i][1];
                int x2 = points[j][0];
                int y2 = points[j][1];

                if (x1 != x2 && y1 != y2) {
                    // Check if the other two corners exist to form a rectangle
                    if (pointSet.contains(x1 + "," + y2) && pointSet.contains(x2 + "," + y1)) {
                        int area = Math.abs(x1 - x2) * Math.abs(y1 - y2);
                        minArea = Math.min(minArea, area);
                    }
                }
            }
        }

        return minArea == Integer.MAX_VALUE ? 0 : minArea;
    }
}

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Problem Info

Description

Problem 939: Minimum Area Rectangle - Detailed explanation and solution approaches.

Topics

Array Hash Table Math Geometry Sorting

Companies

ByteDanceFlipkartSnapVerily

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:939

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

7 steps

1
Step 1: Create a set (or hash table) containing all the given points for efficient lookup.
2
Step 2: Iterate through all possible pairs of points (p1, p2) from the input array.
3
Step 3: For each pair of points (p1(x1, y1), p2(x2, y2)), check if x1 != x2 and y1 != y2. This condition ensures that the two points can be diagonally opposite corners of a rectangle.
4
Step 4: If the condition in Step 3 is met, check if the other two potential corners of the rectangle, (x1, y2) and (x2, y1), are present in the point set.
5
Step 5: If both corners (x1, y2) and (x2, y1) exist in the set, calculate the area of the rectangle as abs(x1 - x2) * abs(y1 - y2).
6
Step 6: Update the minimum area found so far if the calculated area is smaller.
7
Step 7: After iterating through all pairs of points, return the minimum area. If no rectangle was found, return 0.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The nested loops iterate through all possible pairs of points, giving us O(n^2). The hash set stores at most 'n' elements. Therefore, the time and space complexities are O(n^2) and O(n), respectively.

📖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

Sorting Approach

Sort the points based on x-coordinate, then y-coordinate. Iterate through the sorted points, and for each point, check for potential rectangle formation with subsequent points. This approach could potentially prune the search space if we know the expected area is small, however it doesn't change the worst case complexity.

Time: O(n^2) in the worst case, but could be faster on average than the hash set approach depending on the data distribution. Sorting takes O(n log n), and the nested loops would take O(n^2) in the worst case.Space: O(1) if sorting is done in-place, or O(n) if sorting requires auxiliary space.

Trade-offs: The sorting approach might be more efficient on average if the input points are clustered in a way that allows early termination of the search. However, in the worst case, it has the same time complexity as the hash set approach with additional sorting overhead. Use it if there's a high expectation of early termination by pruning the search space due to the distribution of input points.

✨

Key Insights

1Insight 1: To form a rectangle with sides parallel to the axes, we need to find four points such that two pairs of points have the same x-coordinate and the other two pairs have the same y-coordinate.
2Insight 2: Using a set or hash table to store the points allows for efficient checking of whether a potential rectangle's corners exist within the input point set.
3Insight 3: We can iterate through all possible pairs of points and check if they can be two corners of a rectangle. If they are, we check if the other two corners exist in the set of points.

📚

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!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: No rectangle exists. The solution correctly returns 0 in this case, as the minimum area is initialized to infinity (or MAX_VALUE) and remains unchanged if no valid rectangle is found.
Edge case 2: Points are collinear (lie on the same line). The solution correctly handles this case because it requires x1 != x2 and y1 != y2 to form a valid rectangle. If all points are collinear, no valid rectangle will be found, and 0 will be returned.
Edge case 3: Duplicate points in the input. Problem statement specifies that all points are unique. The provided solutions assume unique points and might not produce correct result with duplicate points

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if the other two corners of the rectangle exist. Many beginners might only consider the area formed by two points without verifying the presence of the remaining corners.

✗Common mistake 2: Incorrectly calculating the area or not handling the case where no rectangle exists. Ensure the area calculation is correct using absolute differences in x and y coordinates and properly initialize the minimum area to a large value so that the case of no rectangle being found returns 0.

✗Common mistake 3: Not using a hash set (or other appropriate data structure) to improve the performance of point existence checks. A naive approach might iterate through the entire points array for each point existence check, resulting in O(n^3) time complexity, which will exceed the time limit for larger inputs.

✅

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. Finding rectangular objects in images, such as bounding boxes around detected objects. The algorithm can be used to find the smallest rectangle that encloses a set of feature points.

🏦Real-world scenario 2: Geographic information systems (GIS). Determining the minimum bounding rectangle for a set of geographical points, which can be used to estimate the area covered by a set of locations or objects. This can be used, for example, to determine the minimum area a delivery service would need to consider when planning delivery routes for a customer.

💡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: July 10, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

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"?

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?