# Find the K closest points to origin using Priority Queue

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 ` `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, ` `                   ``vector >, ` `                   ``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); ` ` `  `    ``kClosestPoints(x, y, n, K); ` ` `  `    ``return` `0; ` `} `

Output:

```-2 2
```

Time Complexity: O(n + K * log(n))

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