What is the time complexity of Maximum Square Area by Removing Fences From a Field?

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

What is the space complexity of Maximum Square Area by Removing Fences From a Field?

The space complexity is O(m^2).

What data structures are used to solve Maximum Square Area by Removing Fences From a Field?

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

What are the key insights for solving Maximum Square Area by Removing Fences From a Field?

Insight 1: The problem requires finding equal-length horizontal and vertical gaps within the fences to form a square. Therefore, calculating all possible gaps between fences is crucial. Insight 2: Using a set to store the horizontal gaps allows for efficient checking if a vertical gap of the same length exists.

Which companies ask LeetCode 2975. Maximum Square Area by Removing Fences From a Field in interviews?

This problem is frequently asked at Atlassian.

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Maximum Square Area by Removing Fences From a Field - LeetCode 2975 Solution

Maximum Square Area by Removing Fences From a Field - Complete Solution Guide

Maximum Square Area by Removing Fences From a Field is LeetCode problem 2975, 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

There is a large (m - 1) x (n - 1) rectangular field with corners at (1, 1) and (m, n) containing some horizontal and vertical fences given in arrays hFences and vFences respectively. Horizontal fences are from the coordinates (hFences[i], 1) to (hFences[i], n) and vertical fences are from the coordinates (1, vFences[i]) to (m, vFences[i]) . Return the maximum area of a square field that can be formed by removing some fences ( possibly none ) or -1 if it is impossible to make a square field . Si

Detailed Explanation

The problem asks us to find the maximum area of a square that can be formed within a rectangular field by removing some of the given horizontal and vertical fences. The field is bounded by fences at the edges (1,1) to (1,n), (m,1) to (m,n), (1,1) to (m,1), and (1,n) to (m,n), which cannot be removed. We are given the coordinates of additional horizontal fences (`hFences`) and vertical fences (`vFences`). The goal is to find the largest square area that can be created by selecting a subset of the horizontal and vertical fences, and the surrounding border fences. If no square can be formed, return -1. The area should be returned modulo 10^9 + 7.

Solution Approach

The solution calculates all possible gaps between horizontal fences and stores them in a set. Then, it iterates through all possible gaps between vertical fences and checks if each vertical gap also exists as a horizontal gap. If a matching gap is found, the side length of the square is updated. Finally, the maximum side length is squared (modulo 10^9 + 7) to obtain the maximum square area. If no square can be formed (max_side remains 0), then -1 is returned.

Step-by-Step Algorithm

Step 1: Add the bounding fences (1 and m) to the `hFences` list and (1 and n) to `vFences` list. Convert the lists into arrays if required.
Step 2: Calculate all possible horizontal gaps by iterating through all pairs of horizontal fences and compute their absolute difference. Store these gaps in a set (`h_gaps`).
Step 3: Calculate all possible vertical gaps by iterating through all pairs of vertical fences and compute their absolute difference.
Step 4: For each vertical gap, check if it exists in the `h_gaps` set. If it does, it means we can form a square with that side length.
Step 5: Update the `max_side` with the maximum gap found in both horizontal and vertical direction which are equal. If no equal gaps are found return -1.
Step 6: Calculate the area of the largest square found (max_side * max_side) and return the result modulo 10^9 + 7.

Key Insights

Insight 1: The problem requires finding equal-length horizontal and vertical gaps within the fences to form a square. Therefore, calculating all possible gaps between fences is crucial.
Insight 2: Using a set to store the horizontal gaps allows for efficient checking if a vertical gap of the same length exists.
Insight 3: We need to include the outer boundary fences (at 1 and m/n) in the fence lists to find valid square areas.

Complexity Analysis

Time Complexity: O(m^2 + n^2)

Space Complexity: O(m^2)

Topics

This problem involves: Array, Hash Table, Enumeration.

Companies

Asked at: Atlassian.

Maximum Square Area by Removing Fences From a Field | LeetCode 2975

Medium

Array Hash Table Enumeration

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
    public int maximizeSquareArea(int m, int n, int[] hFences, int[] vFences) {
        int MOD = (int) Math.pow(10, 9) + 7;

        List<Integer> all_hFences = new ArrayList<>();
        for (int fence : hFences) {
            all_hFences.add(fence);
        }
        all_hFences.add(1);
        all_hFences.add(m);

        Set<Integer> h_gaps = new HashSet<>();
        for (int i = 0; i < all_hFences.size(); i++) {
            for (int j = i + 1; j < all_hFences.size(); j++) {
                h_gaps.add(Math.abs(all_hFences.get(i) - all_hFences.get(j)));
            }
        }

        List<Integer> all_vFences = new ArrayList<>();
        for (int fence : vFences) {
            all_vFences.add(fence);
        }
        all_vFences.add(1);
        all_vFences.add(n);

        int max_side = 0;
        for (int i = 0; i < all_vFences.size(); i++) {
            for (int j = i + 1; j < all_vFences.size(); j++) {
                int v_gap = Math.abs(all_vFences.get(i) - all_vFences.get(j));
                if (h_gaps.contains(v_gap)) {
                    max_side = Math.max(max_side, v_gap);
                }
            }
        }

        if (max_side == 0) {
            return -1;
        }

        return (int) (((long) max_side * max_side) % MOD);
    }
}

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Problem Info

Description

Problem 2975: Maximum Square Area by Removing Fences From a Field - Detailed explanation and solution approaches.

Topics

Array Hash Table Enumeration

Companies

Atlassian

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2975

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

6 steps

1
Step 1: Add the bounding fences (1 and m) to the `hFences` list and (1 and n) to `vFences` list. Convert the lists into arrays if required.
2
Step 2: Calculate all possible horizontal gaps by iterating through all pairs of horizontal fences and compute their absolute difference. Store these gaps in a set (`h_gaps`).
3
Step 3: Calculate all possible vertical gaps by iterating through all pairs of vertical fences and compute their absolute difference.
4
Step 4: For each vertical gap, check if it exists in the `h_gaps` set. If it does, it means we can form a square with that side length.
5
Step 5: Update the `max_side` with the maximum gap found in both horizontal and vertical direction which are equal. If no equal gaps are found return -1.
6
Step 6: Calculate the area of the largest square found (max_side * max_side) and return the result modulo 10^9 + 7.

💡

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(m^2 + n^2)

💾

Space Complexity

O(m^2)

📝

Why?

The nested loops for calculating gaps dominate the time complexity. The space complexity is determined by the size of the set used to store horizontal gaps.

📖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 fences and using binary search

Sort both the horizontal and vertical fences and then, when checking if a vertical gap exists as horizontal gap, perform a binary search. This does not necessarily improve performance due to having to iterate through all possible gaps anyway.

Time: O(h log h + v log v + v^2 log h), where h is number of horizontal fences and v is the number of vertical fencesSpace: O(h + v)

Trade-offs: This approach might be beneficial if the number of vertical fences is much smaller than the number of horizontal fences since binary search for vertical gaps in sorted horizontal fences is more efficient than direct lookup in the h_gaps hashset.

✨

Key Insights

1Insight 1: The problem requires finding equal-length horizontal and vertical gaps within the fences to form a square. Therefore, calculating all possible gaps between fences is crucial.
2Insight 2: Using a set to store the horizontal gaps allows for efficient checking if a vertical gap of the same length exists.
3Insight 3: We need to include the outer boundary fences (at 1 and m/n) in the fence lists to find valid square areas.

📚

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: No square can be formed. This happens when there are no common gaps between horizontal and vertical fences. The solution correctly returns -1 in this case.
Edge case 2: m=n and hFences=vFences=[]. The largest possible square will have side m-1=n-1. So the code should return (m-1)*(m-1) % MOD
Edge case 3: Large values of `max_side`. To prevent overflow, the area is calculated modulo (10^9 + 7).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to include the outer boundary fences (1 and m/n) in the calculation of gaps. This can lead to incorrect results if the largest square is formed by the outer fences.

✗Common mistake 2: Integer overflow when calculating the area. The area should be calculated using long long to avoid overflow and then the modulo operation must be applied to get the correct result within the specified range.

✗Common mistake 3: Not handling the case where no square can be formed, leading to an incorrect return value (not -1).

✅

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: Optimizing space utilization in rectangular layouts. For example, finding the largest square-shaped area to place a machine on a factory floor with existing obstacles (fences).

🏦Real-world scenario 2: Resource allocation in a grid-based system. For instance, allocating square-shaped plots of land for development, given pre-existing infrastructure constraints (fences).

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: September 4, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

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

@InterviewProJun 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 "Maximum Square Area by Removing Fences From a Field"?

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?