What is the time complexity of Count Lattice Points Inside a Circle?

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

What is the space complexity of Count Lattice Points Inside a Circle?

The space complexity is O(n*r^2).

What data structures are used to solve Count Lattice Points Inside a Circle?

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

What are the key insights for solving Count Lattice Points Inside a Circle?

Insight 1: Since the constraints on x, y, and r are relatively small (1 <= x, y <= 100, 1 <= r <= min(x, y)), a brute-force approach of iterating through all possible lattice points within a bounding box defined by the circles' coordinates is feasible. Insight 2: Using a set data structure is crucial to efficiently track and count only the *unique* lattice points.

Which companies ask LeetCode 2249. Count Lattice Points Inside a Circle in interviews?

This problem is frequently asked at Rubrik.

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Count Lattice Points Inside a Circle - LeetCode 2249 Solution

Q: How to solve Count Lattice Points Inside a Circle?

The solution iterates through each circle and, for each circle, iterates through all possible lattice points within a square bounding box centered at the circle's center and with side length 2*r. For each point within the bounding box, it checks if the point is inside the circle using the circle equation. If it is, the point is added to a set. Finally, the size of the set is returned, representing the total number of unique lattice points within the circles.

Q: Which companies ask LeetCode 2249. Count Lattice Points Inside a Circle in interviews?

This problem is frequently asked at Rubrik.

Count Lattice Points Inside a Circle - Complete Solution Guide

Count Lattice Points Inside a Circle is LeetCode problem 2249, 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

Given a 2D integer array circles where circles[i] = [x i , y i , r i ] represents the center (x i , y i ) and radius r i of the i th circle drawn on a grid, return the number of lattice points that are present inside at least one circle . Note: A lattice point is a point with integer coordinates. Points that lie on the circumference of a circle are also considered to be inside it. Example 1: Input: circles = [[2,2,1]] Output: 5 Explanation: The figure above shows the given circle. The lattice po

Detailed Explanation

The problem asks us to count the number of unique integer coordinate points (lattice points) that lie inside or on the boundary of at least one circle from a given set of circles. Each circle is defined by its center (x, y) and radius r. The input is a 2D array 'circles' where each row represents a circle [x, y, r]. The output is the total count of distinct lattice points covered by one or more circles.

Solution Approach

The solution iterates through each circle and, for each circle, iterates through all possible lattice points within a square bounding box centered at the circle's center and with side length 2*r. For each point within the bounding box, it checks if the point is inside the circle using the circle equation. If it is, the point is added to a set. Finally, the size of the set is returned, representing the total number of unique lattice points within the circles.

Step-by-Step Algorithm

Step 1: Initialize an empty set called 'points' to store the unique lattice points.
Step 2: Iterate through each circle in the input array 'circles'.
Step 3: For each circle with center (cx, cy) and radius r, iterate through all possible x coordinates from cx - r to cx + r (inclusive).
Step 4: For each x coordinate, iterate through all possible y coordinates from cy - r to cy + r (inclusive).
Step 5: For each point (x, y), check if it lies inside or on the circle using the distance formula (or circle equation): (x - cx)^2 + (y - cy)^2 <= r^2
Step 6: If the point (x, y) lies inside or on the circle, add it to the 'points' set.
Step 7: After iterating through all circles and points, return the size of the 'points' set.

Key Insights

Insight 1: Since the constraints on x, y, and r are relatively small (1 <= x, y <= 100, 1 <= r <= min(x, y)), a brute-force approach of iterating through all possible lattice points within a bounding box defined by the circles' coordinates is feasible.
Insight 2: Using a set data structure is crucial to efficiently track and count only the *unique* lattice points.
Insight 3: The distance formula (or the circle equation) is fundamental to determining if a point lies inside a circle: (x - cx)^2 + (y - cy)^2 <= r^2

Complexity Analysis

Time Complexity: O(n*r^2)

Space Complexity: O(n*r^2)

Topics

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

Companies

Asked at: Rubrik.

Count Lattice Points Inside a Circle | LeetCode 2249

Medium

Array Hash Table Math Geometry Enumeration

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

Solution Code

Available in:

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

class Solution {
    public int countLatticePoints(int[][] circles) {
        Set<String> points = new HashSet<>();
        
        for (int[] circle : circles) {
            int cx = circle[0];
            int cy = circle[1];
            int r = circle[2];
            
            for (int x = cx - r; x <= cx + r; x++) {
                for (int y = cy - r; y <= cy + r; y++) {
                    if ((x - cx) * (x - cx) + (y - cy) * (y - cy) <= r * r) {
                        points.add(x + "," + y);
                    }
                }
            }
        }
        
        return points.size();
    }
}

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Problem Info

Description

Problem 2249: Count Lattice Points Inside a Circle - Detailed explanation and solution approaches.

Topics

Array Hash Table Math Geometry Enumeration

Companies

Rubrik

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2249

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: Initialize an empty set called 'points' to store the unique lattice points.
2
Step 2: Iterate through each circle in the input array 'circles'.
3
Step 3: For each circle with center (cx, cy) and radius r, iterate through all possible x coordinates from cx - r to cx + r (inclusive).
4
Step 4: For each x coordinate, iterate through all possible y coordinates from cy - r to cy + r (inclusive).
5
Step 5: For each point (x, y), check if it lies inside or on the circle using the distance formula (or circle equation): (x - cx)^2 + (y - cy)^2 <= r^2
6
Step 6: If the point (x, y) lies inside or on the circle, add it to the 'points' set.
7
Step 7: After iterating through all circles and points, return the size of the 'points' set.

💡

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*r^2)

💾

Space Complexity

O(n*r^2)

📝

Why?

The time complexity is derived from the nested loops. The outer loop iterates through each circle (O(n)). The inner loops iterate through a square area determined by the radius (O(r*r)). The space complexity comes from storing the unique lattice points in a set. In the worst-case, many unique lattice points can be within each circle, leading to the O(n*r^2) space requirement. The C implementation may have a higher space complexity as the set implemented through a array may dynamically reallocate more space than actually used.

📖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

Sweep Line Algorithm

While less intuitive for this particular problem given the constraints, one could consider a sweep line algorithm. Imagine sweeping a vertical line across the x-axis. For each x-coordinate, determine which circles the line intersects and then iterate through the possible y-coordinates within those circles.

Time: O(N log N + K), where N is number of circles and K is number of lattice points.Space: O(N)

Trade-offs: This approach might offer slight improvements in certain scenarios, especially if the circles are very sparse or clustered, but the overhead of managing the sweep line and intersection calculations makes it less practical than the brute force method given the constraints of this problem. It also requires a better knowledge of geometric algorithms.

✨

Key Insights

1Insight 1: Since the constraints on x, y, and r are relatively small (1 <= x, y <= 100, 1 <= r <= min(x, y)), a brute-force approach of iterating through all possible lattice points within a bounding box defined by the circles' coordinates is feasible.
2Insight 2: Using a set data structure is crucial to efficiently track and count only the *unique* lattice points.
3Insight 3: The distance formula (or the circle equation) is fundamental to determining if a point lies inside a circle: (x - cx)^2 + (y - cy)^2 <= r^2

📚

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: Circles that overlap. The set data structure correctly handles overlapping circles by storing only unique lattice points, preventing double-counting.
Edge case 2: A single very large circle. The algorithm correctly iterates through all possible points within the circle's bounds.
Edge case 3: Multiple circles with very small radii. The algorithm handles this correctly as the inner loops' range adjusts based on the radius.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to use a set or similar data structure to ensure that only unique lattice points are counted, resulting in an incorrect count due to double-counting.

✗Common mistake 2: Incorrectly implementing the distance formula or circle equation, leading to incorrect determination of whether a point lies inside a circle.

✗Common mistake 3: Not handling the boundary condition correctly (i.e., points on the circumference of the circle) by using '<=' instead of '<' in the circle equation check.

✗Common mistake 4: Not correctly calculating the bounds for the inner loops. If cx-r or cy-r are negative, this will cause index out of bounds errors in many languages. This is not an issue here because the problem constraints guarantee x and y >= 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: Coverage analysis for wireless networks. Determining the area covered by multiple Wi-Fi access points or cellular towers, where each access point has a certain range (radius).

🏦Real-world scenario 2: Geographical analysis. Determining the population within a certain radius of multiple points of interest (e.g., hospitals, schools). Each point of interest has a service radius, and the algorithm helps count the number of households within the combined service areas.

🎮Real-world scenario 3: Collision detection in simulations. Checking if objects represented as circles in a 2D space are overlapping or intersecting, which can be applied in game development or physics simulations. The lattice point counting concept could be modified to assess the degree of overlap.

💡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 29, 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 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 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 "Count Lattice Points Inside a Circle"?

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.

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