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++ 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 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 |
# Python3 implementation to find the K # closest points to origin # using Priority Queue import heapq as hq
# Function to find the K closest points def kClosestPoints(x, y, n, k):
# Create a priority queue
pq = []
# Pushing all the points
# in the queue
for i in range (n):
hq.heappush(pq, (x[i] * x[i] + y[i] * y[i],x[i],y[i]))
# Print the first K elements
# of the queue
for i in range (k) :
# Store the top of the queue
# in a temporary pair
p = hq.heappop(pq)
# Print the first (x)
# and second (y) of pair
print ( "{} {}" . format (p[ 1 ],p[ 2 ]))
# Driver code if __name__ = = '__main__' :
# x coordinate of points
x = [ 1 , - 2 ]
# y coordinate of points
y = [ 3 , 2 ]
K = 1
n = len (x)
kClosestPoints(x, y, n, K)
|
// C# implementation to find the K // closest points to origin // using Priority Queue using System;
using System.Collections.Generic;
// Point class to store // a point class Pair : IComparable<Pair> {
public int first, second;
public 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
List<Pair> pq = new List<Pair>();
// Pushing all the points
// in the queue
for ( int i = 0; i < n; i++) {
pq.Add( new Pair(x[i], y[i]));
}
pq.Sort();
// 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[0];
pq.RemoveAt(0);
// Print the first (x)
// and second (y) of pair
Console.WriteLine(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 phasing17 |
<script> // Javascript implementation to find the K // closest points to origin // using Priority Queue // Function to find the K closest points function kClosestPoints(x, y, n, k)
{ // Create a priority queue
var pq = [];
// Pushing all the points
// in the queue
for ( var i = 0; i < n; i++) {
pq.push([x[i], y[i]]);
}
// Print the first K elements
// of the queue
for ( var i = 0; i < k; i++) {
// Store the top of the queue
// in a temporary pair
var p = pq[pq.length-1];
// Print the first (x)
// and second (y) of pair
document.write( p[0] + " "
+ p[1] + "<br>" );
// Remove top element
// of priority queue
pq.pop();
}
} // Driver code // x coordinate of points var x = [1, -2];
// y coordinate of points var y = [3, 2];
var K = 1;
var n = x.length;
kClosestPoints(x, y, n, K); // This code is contributed by rutvik_56. </script> |
-2 2
Time Complexity: O(N + K * log(N))
Auxiliary Space: O(N)