Merge pull request #8 from joney000/priority_queue

add priority queue

Algorithms/PriorityQueue.java

@@ -0,0 +1,36 @@
+class Solution {
+
+    class Point{
+        int x, y;
+        int distance;
+        public Point(int x, int y){
+            this.x = x;
+            this.y = y;
+            this.distance = x *  x + y  * y;
+        }
+
+    }
+
+    // returns the K Closest points from origin (0, 0)
+    // Time: O(n log k), space: O(k)
+    public int[][] kClosest(int[][] points, int k) {
+        if(points.length == 0 || k > points.length){
+            return null;
+        }
+        int numPoints = points.length;
+        PriorityQueue<Point> pQueue = new PriorityQueue<Point>(k + 1, (a,b) -> (b.distance - a.distance)); // max elem on top
+        for(int[] point: points){
+            pQueue.add(new Point(point[0], point[1]));
+            if(pQueue.size() > k){
+                pQueue.poll();
+            }
+        }
+        int[][] sortedElements = new int[k][2];
+        for(int pos = k - 1; pos >= 0; pos--){
+            Point point = (Point)pQueue.poll();
+            sortedElements[pos][0] = point.x;
+            sortedElements[pos][1] = point.y;
+        }
+        return sortedElements;
+    }
+}

README.md

@@ -40,6 +40,7 @@ In This Repository, I have written some of the important Algorithms and Data Str
 | [Kth Order Statics](https://github.com/joney000/Java-Competitive-Programming/blob/master/Algorithms/kthOrderStatics.java)|O(N), O(N) | K’th Smallest/Largest Element in Unsorted Array
 | [Trie / Prefix Tree](https://github.com/joney000/Java-Competitive-Programming/blob/master/Algorithms/Trie.java)| O(N * L), O(N * L)| if there are N strings of L size, per query time(Prefix information) = O(L) 
 | [LIS - Longest Increasing Subsequence](https://github.com/joney000/Java-Competitive-Programming/blob/master/Algorithms/LIS_nLOGn.java)| O(N * log(N)), O(N)
+| [Priority Queue](https://github.com/joney000/Java-Competitive-Programming/blob/master/Algorithms/PriorityQueue.java)| O(log(N)), O(N) | N = no of objects in the queue. peek: O(1), poll/add: O(log n)
 
 ## Contributions