KclosestPointOrigin problem mentioned in 75 Qs

Python/28_K_Closest_Points_To_Orgin/KClosestPointOrigin.py

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+#runtime 653ms. memory 22.21 MB
+import heapq
+
+class Solution:
+    def kClosest(self, points, k):
+        # Let's create a max-heap to store the k closest points based on distance.
+        # We're going to make it max-heap friendly by storing negative distances.
+
+        max_heap = []
+
+        # Now, let's loop through the points and calculate the squared Euclidean distance.
+        # We'll avoid the square root to keep things snappy.
+
+        for point in points:
+            x, y = point
+            distance = x * x + y * y
+
+            # We'll push the distance to the heap with a negative sign.
+            heapq.heappush(max_heap, (-distance, point))
+
+            # If our heap has more than k points, we'll pop the one with the largest distance.
+            if len(max_heap) > k:
+                heapq.heappop(max_heap)
+
+        # Now that we've collected our k closest points, let's reverse the order to get them
+        # in ascending order of distance.
+
+        closest_points = [point for (_, point) in max_heap][::-1]
+
+        # And there you have it! The k closest points.
+
+        return closest_points
+
+# Let's try it out with an example:
+points = [[1, 3], [-2, 2]]
+k = 1
+solution = Solution()
+result = solution.kClosest(points, k)
+print(result)  # Ta-da! The closest point is: [[-2, 2]]