What is the time complexity of K Closest Points to Origin?

The optimal solution has a time complexity of O(nlogk).

What data structures are used to solve K Closest Points to Origin?

This problem uses Array, Math, Divide and Conquer, Geometry, Sorting, Heap (Priority Queue), Quickselect. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving K Closest Points to Origin?

Insight 1: The Euclidean distance calculation involves a square root, but since we only need to compare distances, we can optimize by comparing the squared distances. This avoids the computationally expensive square root operation. Insight 2: A max-heap (priority queue) is well-suited for this problem. It allows us to efficiently keep track of the 'k' closest points seen so far. The max-heap stores the distances, always maintaining the largest distance among the 'k' closest points at the top.

Which companies ask LeetCode 973. K Closest Points to Origin in interviews?

This problem is frequently asked at Asana, Axon, Snap, Swiggy, Visa and 1 more companies.

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K Closest Points to Origin - LeetCode 973 Solution

K Closest Points to Origin - Complete Solution Guide

K Closest Points to Origin is LeetCode problem 973, 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 an array of points where points[i] = [x i , y i ] represents a point on the X-Y plane and an integer k , return the k closest points to the origin (0, 0) . The distance between two points on the X-Y plane is the Euclidean distance (i.e., √(x 1 - x 2 ) 2 + (y 1 - y 2 ) 2 ). You may return the answer in any order . The answer is guaranteed to be unique (except for the order that it is in). Example 1: Input: points = [[1,3],[-2,2]], k = 1 Output: [[-2,2]] Explanation: The distance betwe

Detailed Explanation

The problem asks us to find the 'k' points in a given array of points that are closest to the origin (0, 0) in a 2D plane. The distance between two points is calculated using the Euclidean distance formula. The output should be a list containing these 'k' closest points. The order of the points in the output doesn't matter, and the solution is guaranteed to be unique (except for the order).

Solution Approach

The provided solution utilizes a max-heap (priority queue) to keep track of the 'k' closest points encountered so far. The algorithm iterates through each point in the input array. For each point, it calculates the squared Euclidean distance from the origin. If the heap has fewer than 'k' points, the current point is added to the heap. Otherwise, if the current point's squared distance is less than the squared distance of the farthest point currently in the heap (the root of the max-heap), the farthest point is removed, and the current point is added. After processing all points, the heap contains the 'k' closest points to the origin, which are then returned as the result.

Step-by-Step Algorithm

Step 1: Initialize an empty max-heap (priority queue) to store the 'k' closest points and their distances.
Step 2: Iterate through the input array of points.
Step 3: For each point, calculate its squared Euclidean distance from the origin.
Step 4: If the size of the heap is less than 'k', add the point and its squared distance (negated) to the heap. Negation ensures it behaves as max-heap for smallest distances.
Step 5: If the size of the heap is equal to 'k', compare the squared distance of the current point with the largest squared distance in the heap (the root element).
Step 6: If the current point's squared distance is smaller than the largest squared distance in the heap, remove the root of the heap and add the current point and its (negated) squared distance to the heap.
Step 7: After iterating through all the points, the heap will contain the 'k' closest points. Extract these points from the heap and return them as the result.

Key Insights

Insight 1: The Euclidean distance calculation involves a square root, but since we only need to compare distances, we can optimize by comparing the squared distances. This avoids the computationally expensive square root operation.
Insight 2: A max-heap (priority queue) is well-suited for this problem. It allows us to efficiently keep track of the 'k' closest points seen so far. The max-heap stores the distances, always maintaining the largest distance among the 'k' closest points at the top.
Insight 3: We can iterate through the points. If the current point's distance is less than the maximum distance currently in our max-heap, we can replace the farthest point in the max-heap with the current point.

Complexity Analysis

Time Complexity: O(nlogk)

Space Complexity: O(k)

Topics

This problem involves: Array, Math, Divide and Conquer, Geometry, Sorting, Heap (Priority Queue), Quickselect.

Companies

Asked at: Asana, Axon, Snap, Swiggy, Visa, Wix.

K Closest Points to Origin | LeetCode 973

Medium

Array Math Divide and Conquer Geometry Sorting Heap (Priority Queue)Quickselect

Time: O(nlogk)Space: O(k)

Solution Code

Available in:

import java.util.PriorityQueue;
import java.util.Arrays;

class Solution {
    public int[][] kClosest(int[][] points, int k) {
        PriorityQueue<int[]> maxHeap = new PriorityQueue<>((a, b) -> (b[0] * b[0] + b[1] * b[1]) - (a[0] * a[0] + a[1] * a[1]));
        
        for (int[] point : points) {
            
            if (maxHeap.size() < k) {
                maxHeap.offer(point);
            } else {
                int[] farthest = maxHeap.peek();
                if ((point[0] * point[0] + point[1] * point[1]) < (farthest[0] * farthest[0] + farthest[1] * farthest[1])) {
                    maxHeap.poll();
                    maxHeap.offer(point);
                }
            }
        }

        int[][] result = new int[k][2];
        int i = 0;
        for (int[] point : maxHeap) {
            result[i++] = point;
        }
        
        return result;
    }
}

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Problem Info

Description

Problem 973: K Closest Points to Origin - Detailed explanation and solution approaches.

Topics

Array Math Divide and Conquer Geometry Sorting Heap (Priority Queue)Quickselect

Companies

AsanaAxonSnapSwiggyVisaWix

Stats

Difficulty:Medium

Languages:4

Companies:6

Problem #:973

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 max-heap (priority queue) to store the 'k' closest points and their distances.
2
Step 2: Iterate through the input array of points.
3
Step 3: For each point, calculate its squared Euclidean distance from the origin.
4
Step 4: If the size of the heap is less than 'k', add the point and its squared distance (negated) to the heap. Negation ensures it behaves as max-heap for smallest distances.
5
Step 5: If the size of the heap is equal to 'k', compare the squared distance of the current point with the largest squared distance in the heap (the root element).
6
Step 6: If the current point's squared distance is smaller than the largest squared distance in the heap, remove the root of the heap and add the current point and its (negated) squared distance to the heap.
7
Step 7: After iterating through all the points, the heap will contain the 'k' closest points. Extract these points from the heap and return them as the result.

💡

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(nlogk)

💾

Space Complexity

O(k)

📝

Why?

The time complexity is determined by the 'n' iterations through the points array, and the O(logk) heap operations inside the loop. The space complexity depends on the size of the heap which can grow up to 'k' 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

Sorting

Sort all the points based on their distance from the origin and then take the first 'k' elements.

Time: O(nlogn), where n is the number of points.Space: O(n) for creating a copy of the array or modifying the array in-place.

Trade-offs: Sorting has a higher time complexity than the heap-based approach (O(nlogn) vs O(nlogk)). The heap-based approach is better when k is much smaller than n. Sorting is simpler to implement if in-place sorting is acceptable, but modifying the original data may not always be desired.

Quickselect

Use the Quickselect algorithm (similar to Quicksort's partitioning) to find the kth smallest distance. Then, iterate through the array and collect all points with distances less than or equal to the kth smallest distance.

Time: O(n) on average, O(n^2) in the worst case, where n is the number of points.Space: O(1) if done in-place, O(n) otherwise.

Trade-offs: Quickselect has a better average time complexity O(n), compared to O(nlogk) with heap and O(nlogn) with sorting. However it has a worst-case time complexity of O(n^2). Additionally, implementations can be complex. It is preferred when one needs optimal average case performance and is ready to handle possible edge cases carefully.

✨

Key Insights

1Insight 1: The Euclidean distance calculation involves a square root, but since we only need to compare distances, we can optimize by comparing the squared distances. This avoids the computationally expensive square root operation.
2Insight 2: A max-heap (priority queue) is well-suited for this problem. It allows us to efficiently keep track of the 'k' closest points seen so far. The max-heap stores the distances, always maintaining the largest distance among the 'k' closest points at the top.
3Insight 3: We can iterate through the points. If the current point's distance is less than the maximum distance currently in our max-heap, we can replace the farthest point in the max-heap with the current point.

📚

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: When 'k' is equal to the number of points. In this case, the algorithm will add all points to the heap. This case is handled correctly as it is just a normal execution of the algorithm.
Edge case 2: When all points are the same and coincident with the origin. The algorithm correctly returns any k points because the distances will all be zero, hence any k would be closest.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to use the squared distance instead of the actual Euclidean distance. This can lead to unnecessary square root calculations, impacting performance.

✗Common mistake 2: Not using a max-heap, but a min-heap instead. This makes it difficult to efficiently track the farthest point among the 'k' closest points encountered so far.

✅

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: Location-based services (e.g., finding the 'k' nearest restaurants to a user's current location).

🏦Real-world scenario 2: Recommendation systems (e.g., finding the 'k' most similar users to a given user based on their preferences).

🎮Real-world scenario 3: Anomaly detection (e.g., in a dataset of sensor readings, find the 'k' sensors with readings that are most different from the norm, indicating a potential malfunction).

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 13, 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 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 "K Closest Points to Origin"?

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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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