What is the time complexity of Maximum Area Rectangle With Point Constraints II?

The optimal solution has a time complexity of O(n*log(n)).

What data structures are used to solve Maximum Area Rectangle With Point Constraints II?

This problem uses Array, Math, Binary Indexed Tree, Segment Tree, Geometry, Sorting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Area Rectangle With Point Constraints II?

Insight 1: The problem can be solved by iterating through pairs of x-coordinates within each horizontal line (same y-coordinate). Insight 2: Using a segment tree with lazy propagation allows for efficient querying of the highest y-coordinate between a given range of x-coordinates.

Which companies ask LeetCode 3382. Maximum Area Rectangle With Point Constraints II in interviews?

This problem is frequently asked at UKG.

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Maximum Area Rectangle With Point Constraints II - LeetCode 3382 Solution

Maximum Area Rectangle With Point Constraints II - Complete Solution Guide

Maximum Area Rectangle With Point Constraints II is LeetCode problem 3382, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There are n points on an infinite plane. You are given two integer arrays xCoord and yCoord where (xCoord[i], yCoord[i]) represents the coordinates of the i th point. Your task is to find the maximum area of a rectangle that: Can be formed using four of these points as its corners. Does not contain any other point inside or on its border. Has its edges parallel to the axes. Return the maximum area that you can obtain or -1 if no such rectangle is possible. Example 1: Input: xCoord = [1,1,3,3], y

Detailed Explanation

The problem asks us to find the maximum area of a rectangle formed by four points from a given set of points on a 2D plane, such that the rectangle's sides are parallel to the axes and no other points from the set lie inside or on the border of the rectangle. If no such rectangle exists, we should return -1.

Solution Approach

The solution uses a segment tree with lazy propagation to efficiently find the maximum area rectangle. It first groups points by their y-coordinates. Then, it iterates through the y-coordinates in sorted order. For each y-coordinate, it considers all pairs of x-coordinates on that line. It uses the segment tree to find the highest y-coordinate seen so far between the two x-coordinates. If such a y-coordinate exists, a potential rectangle has been identified, and the area is calculated. After considering all pairs of x-coordinates for a given y, the solution updates the segment tree with the current y-coordinate between each adjacent x-coordinates to mark it as the largest y coordinate up to that point.

Step-by-Step Algorithm

Step 1: Preprocess the input by extracting unique x-coordinates and creating a mapping from x-coordinate value to its index in the sorted unique x-coordinates list.
Step 2: Group the x-coordinates based on their corresponding y-coordinates.
Step 3: Sort the y-coordinates to process them in ascending order.
Step 4: Initialize a segment tree with lazy propagation. The segment tree is used to store the highest y-coordinate encountered for each range of x-coordinate indices.
Step 5: Iterate through the sorted y-coordinates.
Step 6: For each y-coordinate, iterate through the x-coordinates associated with that y-coordinate. Consider pairs of adjacent x-coordinates.
Step 7: Query the segment tree for the maximum y-coordinate present between the indices of the two x-coordinates.
Step 8: If a valid y-coordinate is returned from the segment tree (i.e., a rectangle can be formed), calculate the area of the rectangle and update the maximum area if necessary.
Step 9: After processing all x-coordinate pairs for the current y-coordinate, update the segment tree with the current y-coordinate for the ranges between adjacent x-coordinate indices.
Step 10: Return the maximum area found, or -1 if no valid rectangle was found.

Key Insights

Insight 1: The problem can be solved by iterating through pairs of x-coordinates within each horizontal line (same y-coordinate).
Insight 2: Using a segment tree with lazy propagation allows for efficient querying of the highest y-coordinate between a given range of x-coordinates.
Insight 3: Sorting the y-coordinates allows us to process the problem in increasing order of y values, enabling the segment tree to track the highest previously encountered y-coordinate for a given x-range.

Complexity Analysis

Time Complexity: O(n*log(n))

Space Complexity: O(n)

Topics

This problem involves: Array, Math, Binary Indexed Tree, Segment Tree, Geometry, Sorting.

Companies

Asked at: UKG.

Maximum Area Rectangle With Point Constraints II | LeetCode 3382

Hard

Array Math Binary Indexed Tree Segment Tree Geometry Sorting

Time: O(n*log(n))Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public long maxRectangleArea(int[] xCoord, int[] yCoord) {
        int n = xCoord.length;
        if (n < 4) {
            return -1;
        }

        Set<Integer> xSet = new HashSet<>();
        for (int x : xCoord) {
            xSet.add(x);
        }

        List<Integer> all_x = new ArrayList<>(xSet);
        Collections.sort(all_x);

        Map<Integer, Integer> x_map = new HashMap<>();
        for (int i = 0; i < all_x.size(); i++) {
            x_map.put(all_x.get(i), i);
        }

        int num_unique_x = all_x.size();

        if (num_unique_x < 2) {
            return -1;
        }

        Map<Integer, List<Integer>> y_groups = new HashMap<>();
        for (int i = 0; i < n; i++) {
            if (!y_groups.containsKey(yCoord[i])) {
                y_groups.put(yCoord[i], new ArrayList<>());
            }
            y_groups.get(yCoord[i]).add(xCoord[i]);
        }

        for (List<Integer> y_group : y_groups.values()) {
            Collections.sort(y_group);
        }
        
        List<Integer> sorted_y = new ArrayList<>(y_groups.keySet());
        Collections.sort(sorted_y);
        
        long max_area = 0;
        
        int tree_size = num_unique_x - 1;
        long[] tree = new long[4 * tree_size];
        long[] lazy = new long[4 * tree_size];
        Arrays.fill(tree, -1);
        Arrays.fill(lazy, -1);

        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        

        for (int y : sorted_y) {
            List<Integer> x_coords = y_groups.get(y);
            for (int i = 0; i < x_coords.size() - 1; i++) {
                int x1 = x_coords.get(i);
                int x2 = x_coords.get(i+1);
                int idx1 = x_map.get(x1);
                int idx2 = x_map.get(x2);
                if (idx1 >= idx2) continue;

                long y_prev = query(0, 0, tree_size - 1, idx1, idx2 - 1, tree, lazy);
                if (y_prev != -1) {
                    long area = (long)(x2 - x1) * (y - y_prev);
                    if (area > max_area) {
                        max_area = area;
                    }
                }
            }

            for (int i = 0; i < x_coords.size() - 1; i++) {
                int x1 = x_coords.get(i);
                int x2 = x_coords.get(i+1);
                int idx1 = x_map.get(x1);
                int idx2 = x_map.get(x2);
                if (idx1 >= idx2) continue;
                
                update(0, 0, tree_size - 1, idx1, idx2 - 1, y, tree, lazy);
            }
        }

        return max_area > 0 ? max_area : -1;
    }
    
    private void push(int node, int start, int end, long[] tree, long[] lazy) {
        if (lazy[node] != -1) {
            tree[node] = lazy[node];
            if (start != end) {
                lazy[2 * node + 1] = lazy[node];
                lazy[2 * node + 2] = lazy[node];
            }
            lazy[node] = -1;
        }
    }
    
    private void update(int node, int start, int end, int l, int r, int val, long[] tree, long[] lazy) {
        push(node, start, end, tree, lazy);
        if (start > end || start > r || end < l) {
            return;
        }

        if (l <= start && end <= r) {
            lazy[node] = val;
            push(node, start, end, tree, lazy);
            return;
        }

        int mid = (start + end) / 2;
        update(2 * node + 1, start, mid, l, r, val, tree, lazy);
        update(2 * node + 2, mid + 1, end, l, r, val, tree, lazy);
        push(2*node+1, start, mid, tree, lazy);
        push(2*node+2, mid+1, end, tree, lazy);
        tree[node] = Math.max(tree[2 * node + 1], tree[2 * node + 2]);
    }

    private long query(int node, int start, int end, int l, int r, long[] tree, long[] lazy) {
        if (start > end || start > r || end < l) {
            return -1;
        }

        push(node, start, end, tree, lazy);
        if (l <= start && end <= r) {
            return tree[node];
        }

        int mid = (start + end) / 2;
        long p1 = query(2 * node + 1, start, mid, l, r, tree, lazy);
        long p2 = query(2 * node + 2, mid + 1, end, l, r, tree, lazy);
        return Math.max(p1, p2);
    }
}

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Problem Info

Description

Problem 3382: Maximum Area Rectangle With Point Constraints II - Detailed explanation and solution approaches.

Topics

Array Math Binary Indexed Tree Segment Tree Geometry Sorting

Companies

UKG

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3382

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: Preprocess the input by extracting unique x-coordinates and creating a mapping from x-coordinate value to its index in the sorted unique x-coordinates list.
2
Step 2: Group the x-coordinates based on their corresponding y-coordinates.
3
Step 3: Sort the y-coordinates to process them in ascending order.
4
Step 4: Initialize a segment tree with lazy propagation. The segment tree is used to store the highest y-coordinate encountered for each range of x-coordinate indices.
5
Step 5: Iterate through the sorted y-coordinates.
6
Step 6: For each y-coordinate, iterate through the x-coordinates associated with that y-coordinate. Consider pairs of adjacent x-coordinates.
7
Step 7: Query the segment tree for the maximum y-coordinate present between the indices of the two x-coordinates.
8
Step 8: If a valid y-coordinate is returned from the segment tree (i.e., a rectangle can be formed), calculate the area of the rectangle and update the maximum area if necessary.
9
Step 9: After processing all x-coordinate pairs for the current y-coordinate, update the segment tree with the current y-coordinate for the ranges between adjacent x-coordinate indices.
10
Step 10: Return the maximum area found, or -1 if no valid rectangle was found.

💡

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*log(n))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant factor in time complexity is the segment tree operations which take O(log n) time each and these are performed within a loop that iterates up to n times. The space complexity is dominated by the maps and the segment tree which use space proportional to the number of points.

📖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

Iterate through all possible combinations of four points and check if they form a valid rectangle with no points inside. Also ensure all sides are parallel to the axis.

Time: O(n^5), where n is the number of points (O(n^4) for iterating through combinations, O(n) to check if there are points inside the rectangle).Space: O(1)

Trade-offs: This approach is straightforward to implement but extremely inefficient for large datasets. It's only suitable for very small input sizes.

✨

Key Insights

1Insight 1: The problem can be solved by iterating through pairs of x-coordinates within each horizontal line (same y-coordinate).
2Insight 2: Using a segment tree with lazy propagation allows for efficient querying of the highest y-coordinate between a given range of x-coordinates.
3Insight 3: Sorting the y-coordinates allows us to process the problem in increasing order of y values, enabling the segment tree to track the highest previously encountered y-coordinate for a given x-range.

📚

Key Terms Explained

For Students

Array

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

Sorting

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

Edge Cases

Edge case 1: Fewer than four points exist. The solution returns -1, as it's impossible to form a rectangle.
Edge case 2: All points lie on the same line (either horizontal or vertical). The solution returns -1, since a rectangle cannot be formed.
Edge case 3: Overlapping points. The description says all points are unique so overlapping points will not exist.
Edge case 4: No valid rectangle can be formed because all possible rectangles enclose other points. The algorithm returns -1.
Edge case 5: Zero area rectangles due to equal x or y coordinates for adjacent points will be avoided because x_map will skip identical x's

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where fewer than four points are given.

✗Common mistake 2: Incorrectly calculating the area of the rectangle.

✗Common mistake 3: Failing to check if other points lie inside the rectangle.

✗Common mistake 4: Using inefficient data structures or algorithms that result in exceeding the time limit.

✗Common mistake 5: Forgetting to sort the y-coordinates, which is crucial for the segment tree approach to work correctly.

✅

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: Finding the largest rectangular area in a city planning problem with building locations as points, with constraints on building density within that area.

🏦Real-world scenario 2: Optimizing resource allocation by finding the largest rectangular region in a grid where resources are sparsely distributed, avoiding regions with high concentration of existing resources.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: October 21, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

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

@CodeNinjaJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Area Rectangle With Point Constraints II"?

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

🤖 Need help with this problem?