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 |
-2 2
Time Complexity: O(n + K * log(n))
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