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Find the K closest points to origin using Priority Queue

  • Difficulty Level : Medium
  • Last Updated : 14 Aug, 2021

Given a list of n points on 2D plane, the task is to find the K (k < n) closest points to the origin O(0, 0). 
Note: The distance between a point P(x, y) and O(0, 0) using the standard Euclidean Distance
Examples:

Input: [(1, 0), (2, 1), (3, 6), (-5, 2), (1, -4)], K = 3 
Output: [(1, 0), (2, 1), (1, -4)] 
Explanation: 
Square of Distances of points from origin are 
(1, 0) : 1 
(2, 1) : 5 
(3, 6) : 45 
(-5, 2) : 29 
(1, -4) : 17 
Hence for K = 3, the closest 3 points are (1, 0), (2, 1) & (1, -4).
Input: [(1, 3), (-2, 2)], K = 1 
Output: [(-2, 2)] 
Explanation: 
Square of Distances of points from origin are 
(1, 3) : 10 
(-2, 2) : 8 
Hence for K = 1, the closest point is (-2, 2).

Approach using sorting based on distance: This approach is explained in this article.
Approach using Priority Queue for comparison: To solve the problem mentioned above, the main idea is to store the coordinates of the point in a priority queue of pairs, according to the distance of the point from the origin. For assigning the maximum priority to the least distant point from the origin, we use the Comparator class in Priority Queue. Then print the first K elements of the priority queue.
Below is the implementation of above approach:

C++




// C++ implementation to find the K
// closest points to origin
// using Priority Queue
 
#include <bits/stdc++.h>
using namespace std;
 
// Comparator class to assign
// priority to coordinates
class comp {
 
public:
    bool operator()(pair<int, int> a,
                    pair<int, int> b)
    {
        int x1 = a.first * a.first;
        int y1 = a.second * a.second;
        int x2 = b.first * b.first;
        int y2 = b.second * b.second;
 
        // return true if distance
        // of point 1 from origin
        // is greater than distance of
        // point 2 from origin
        return (x1 + y1) > (x2 + y2);
    }
};
 
// Function to find the K closest points
void kClosestPoints(int x[], int y[],
                    int n, int k)
{
    // Create a priority queue
    priority_queue<pair<int, int>,
                   vector<pair<int, int> >,
                   comp>
        pq;
 
    // Pushing all the points
    // in the queue
    for (int i = 0; i < n; i++) {
        pq.push(make_pair(x[i], y[i]));
    }
 
    // Print the first K elements
    // of the queue
    for (int i = 0; i < k; i++) {
 
        // Store the top of the queue
        // in a temporary pair
        pair<int, int> p = pq.top();
 
        // Print the first (x)
        // and second (y) of pair
        cout << p.first << " "
             << p.second << endl;
 
        // Remove top element
        // of priority queue
        pq.pop();
    }
}
 
// Driver code
int main()
{
    // x coordinate of points
    int x[] = { 1, -2 };
 
    // y coordinate of points
    int y[] = { 3, 2 };
    int K = 1;
 
    int n = sizeof(x) / sizeof(x[0]);
 
    kClosestPoints(x, y, n, K);
 
    return 0;
}

Java




// Java implementation to find the K
// closest points to origin
// using Priority Queue
import java.util.*;
 
// Point class to store
// a point
class Pair implements Comparable<Pair>
{
    int first, second;
    Pair(int a, int b)
    {
        first = a;
        second = b;
    }
     
    public int compareTo(Pair o)
    {
        int x1 = first * first;
        int y1 = second * second;
        int x2 = o.first * o.first;
        int y2 = o.second * o.second;
        return (x1 + y1) - (x2 + y2);
    }
}
 
class GFG{
     
// Function to find the K closest points
static void kClosestPoints(int x[], int y[],
                           int n, int k)
{
    // Create a priority queue
    PriorityQueue<Pair> pq = new PriorityQueue<>();
 
    // Pushing all the points
    // in the queue
    for(int i = 0; i < n; i++)
    {
        pq.add(new Pair(x[i], y[i]));
    }
 
    // Print the first K elements
    // of the queue
    for(int i = 0; i < k; i++)
    {
 
        // Remove the top of the queue
        // and store in a temporary pair
        Pair p = pq.poll();
 
        // Print the first (x)
        // and second (y) of pair
        System.out.println(p.first +
                     " " + p.second);
    }
}
 
// Driver code
public static void main(String[] args)
{
     
    // x coordinate of points
    int x[] = { 1, -2 };
 
    // y coordinate of points
    int y[] = { 3, 2 };
    int K = 1;
 
    int n = x.length;
 
    kClosestPoints(x, y, n, K);
}
}
 
// This code is contributed by jrishabh99
Output: 
-2 2

 

Time Complexity: O(N + K * log(N))
Auxiliary Space: O(N)
 

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