What is the time complexity of Maximum Points Inside the Square?

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

What data structures are used to solve Maximum Points Inside the Square?

This problem uses Array, Hash Table, String, Binary Search, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Points Inside the Square?

Insight 1: The distance of a point from the origin determines the minimum size square required to enclose that point. This distance can be calculated as max(abs(x), abs(y)). Insight 2: Sorting points by their distance from the origin allows us to incrementally expand the square and check for tag conflicts.

Which companies ask LeetCode 3143. Maximum Points Inside the Square in interviews?

This problem is frequently asked at HashedIn.

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Maximum Points Inside the Square - LeetCode 3143 Solution

Maximum Points Inside the Square - Complete Solution Guide

Maximum Points Inside the Square is LeetCode problem 3143, 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 a 2D array points and a string s where, points[i] represents the coordinates of point i , and s[i] represents the tag of point i . A valid square is a square centered at the origin (0, 0) , has edges parallel to the axes, and does not contain two points with the same tag. Return the maximum number of points contained in a valid square. Note: A point is considered to be inside the square if it lies on or within the square's boundaries. The side length of the square can be zero. Exam

Detailed Explanation

The problem requires finding the maximum number of points that can be contained within a square centered at the origin (0, 0) with sides parallel to the axes. A valid square is one where no two points within the square have the same tag (character from the input string 's'). The side length of the square can be zero, and points on the boundary are considered inside. The input consists of a 2D array `points` representing point coordinates and a string `s` representing the tag for each point. The output is the maximum number of points in a valid square.

Solution Approach

The solution first calculates the distance of each point from the origin using max(abs(x), abs(y)). Then, it sorts the points based on their distance from the origin. The algorithm iterates through the sorted points, expanding the square size incrementally. In each iteration, it checks if adding the points at the current distance would violate the tag constraint. If it does, the algorithm returns the current count. If not, it updates the count and the set of seen tags.

Step-by-Step Algorithm

Step 1: Create a list of tuples/pairs containing the distance of each point from the origin and its corresponding tag.
Step 2: Sort the list of tuples/pairs by distance in ascending order.
Step 3: Initialize a set `seen_tags` to keep track of the tags seen so far and a variable `count` to store the number of points inside the valid square.
Step 4: Iterate through the sorted list of tuples/pairs.
Step 5: In each iteration, identify the group of points with the same distance from the origin.
Step 6: For each group of points with the same distance, check if adding any of their tags to the `seen_tags` set would result in a conflict (duplicate tag). If there is a conflict, return the current `count`.
Step 7: If there is no conflict, add all the tags from the current group to the `seen_tags` set and increment the `count` by the number of points in the current group.
Step 8: Continue iterating until all points have been processed.
Step 9: Return the final `count`.

Key Insights

Insight 1: The distance of a point from the origin determines the minimum size square required to enclose that point. This distance can be calculated as max(abs(x), abs(y)).
Insight 2: Sorting points by their distance from the origin allows us to incrementally expand the square and check for tag conflicts.
Insight 3: Using a set to keep track of seen tags helps efficiently detect tag conflicts as the square expands. If a duplicate tag is found, the current count is the maximum possible and the algorithm should terminate.
Insight 4: We need to check for tag conflicts *within the points at the current distance*. Therefore, we can use a temporary set to hold these, and *then* add to the larger `seen_tags` set.

Complexity Analysis

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

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, String, Binary Search, Sorting.

Companies

Asked at: HashedIn.

Maximum Points Inside the Square | LeetCode 3143

Medium

Array Hash Table String Binary Search Sorting

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

Solution Code

Available in:

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

class Solution {
    public int maxPointsInsideSquare(int[][] points, String s) {
        int n = points.length;

        List<Pair<Integer, Character>> dist_tags = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            int dist = Math.max(Math.abs(points[i][0]), Math.abs(points[i][1]));
            dist_tags.add(new Pair<>(dist, s.charAt(i)));
        }

        dist_tags.sort((a, b) -> a.getKey().compareTo(b.getKey()));

        Set<Character> seen_tags = new HashSet<>();
        int count = 0;
        int i = 0;
        while (i < n) {
            int current_dist = dist_tags.get(i).getKey();

            int j = i;
            while (j < n && dist_tags.get(j).getKey() == current_dist) {
                j++;
            }

            List<Character> group_tags_list = new ArrayList<>();
            for (int k = i; k < j; k++) {
                group_tags_list.add(dist_tags.get(k).getValue());
            }

            Set<Character> temp_group_tags_set = new HashSet<>();
            for (Character tag : group_tags_list) {
                if (seen_tags.contains(tag) || temp_group_tags_set.contains(tag)) {
                    return count;
                }
                temp_group_tags_set.add(tag);
            }

            count += group_tags_list.size();
            seen_tags.addAll(temp_group_tags_set);

            i = j;

        return count;
    }

    // Helper class to represent a pair of values
    private static class Pair<K, V> {
        private final K key;
        private final V value;

        public Pair(K key, V value) {
            this.key = key;
            this.value = value;
        }

        public K getKey() {
            return key;
        }

        public V getValue() {
            return value;
        }
    }
}

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Problem Info

Description

Problem 3143: Maximum Points Inside the Square - Detailed explanation and solution approaches.

Topics

Array Hash Table String Binary Search Sorting

Companies

HashedIn

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3143

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

9 steps

1
Step 1: Create a list of tuples/pairs containing the distance of each point from the origin and its corresponding tag.
2
Step 2: Sort the list of tuples/pairs by distance in ascending order.
3
Step 3: Initialize a set `seen_tags` to keep track of the tags seen so far and a variable `count` to store the number of points inside the valid square.
4
Step 4: Iterate through the sorted list of tuples/pairs.
5
Step 5: In each iteration, identify the group of points with the same distance from the origin.
6
Step 6: For each group of points with the same distance, check if adding any of their tags to the `seen_tags` set would result in a conflict (duplicate tag). If there is a conflict, return the current `count`.
7
Step 7: If there is no conflict, add all the tags from the current group to the `seen_tags` set and increment the `count` by the number of points in the current group.
8
Step 8: Continue iterating until all points have been processed.
9
Step 9: Return the final `count`.

💡

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 time complexity is dominated by the sorting algorithm. The sorting takes O(n log n) time. The rest of the operations (calculating distances, set lookups and insertions) take O(n) time in total on average, but this is dominated by the sorting. The space complexity comes from the `dist_tags` array (or similar data structure) and the `seen_tags` set, both potentially storing n elements.

📖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 square sizes and for each square, iterate through all points to check if they are inside the square and if there are any tag conflicts. This approach is extremely inefficient.

Time: O(n^3) or higher. Checking all square sizes is inefficient. For each size, checking all tags takes O(n^2) if you check all pairs for duplicates each time.Space: O(n) to store the points inside the square for each iteration.

Trade-offs: The brute force approach is simpler to understand but has a very high time complexity, making it unsuitable for larger datasets. The main solution is much more efficient.

✨

Key Insights

1Insight 1: The distance of a point from the origin determines the minimum size square required to enclose that point. This distance can be calculated as max(abs(x), abs(y)).
2Insight 2: Sorting points by their distance from the origin allows us to incrementally expand the square and check for tag conflicts.
3Insight 3: Using a set to keep track of seen tags helps efficiently detect tag conflicts as the square expands. If a duplicate tag is found, the current count is the maximum possible and the algorithm should terminate.
4Insight 4: We need to check for tag conflicts *within the points at the current distance*. Therefore, we can use a temporary set to hold these, and *then* add to the larger `seen_tags` set.

📚

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!

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Sorting

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

Edge Cases

Edge case 1: All points have the same tag. The algorithm will correctly return 0 since the first tag will be added, and the second will cause the function to return.
Edge case 2: No points. The algorithm will correctly return 0 since the loop will not execute.
Edge case 3: Points with zero distance (origin). The algorithm handles this correctly because those points are considered inside the square if no tag conflict arises.
Edge case 4: Large coordinates. The algorithm uses `max(abs(x), abs(y))` to calculate the distances which works correctly for large coordinates.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the points by distance. Without sorting, it's difficult to incrementally build valid squares.

✗Common mistake 2: Incorrectly calculating the distance from the origin. Using Euclidean distance (sqrt(x^2 + y^2)) instead of max(abs(x), abs(y)) would lead to incorrect results.

✗Common mistake 3: Failing to handle the case where two points at the same distance have the same tag.

✗Common mistake 4: Not using a temporary set to check for conflicts within a group of points with the same distance, but instead only checking against the global `seen_tags` set, which will lead to incorrect results because you will add a point that makes other points at same distance invalid before it is too late.

✅

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 resource allocation in a grid-based system, where each location has a certain attribute and the objective is to maximize resource usage within a defined area without conflicting attributes.

🏦Real-world scenario 2: Image processing, where points represent pixels and tags represent color values. The problem could be related to identifying the largest homogeneous region centered in the image, subject to constraints on color variation.

💡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.

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 23, 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 13

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

@TechLeadJul 6

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

@DevExpert92Jun 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 Points Inside the Square"?

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?