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Author: jimit105

1014-k-closest-points-to-origin/README.md

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+<p>Given an array of <code>points</code> where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> represents a point on the <strong>X-Y</strong> plane and an integer <code>k</code>, return the <code>k</code> closest points to the origin <code>(0, 0)</code>.</p>
+
+<p>The distance between two points on the <strong>X-Y</strong> plane is the Euclidean distance (i.e., <code>&radic;(x<sub>1</sub> - x<sub>2</sub>)<sup>2</sup> + (y<sub>1</sub> - y<sub>2</sub>)<sup>2</sup></code>).</p>
+
+<p>You may return the answer in <strong>any order</strong>. The answer is <strong>guaranteed</strong> to be <strong>unique</strong> (except for the order that it is in).</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/03/03/closestplane1.jpg" style="width: 400px; height: 400px;" />
+<pre>
+<strong>Input:</strong> points = [[1,3],[-2,2]], k = 1
+<strong>Output:</strong> [[-2,2]]
+<strong>Explanation:</strong>
+The distance between (1, 3) and the origin is sqrt(10).
+The distance between (-2, 2) and the origin is sqrt(8).
+Since sqrt(8) &lt; sqrt(10), (-2, 2) is closer to the origin.
+We only want the closest k = 1 points from the origin, so the answer is just [[-2,2]].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> points = [[3,3],[5,-1],[-2,4]], k = 2
+<strong>Output:</strong> [[3,3],[-2,4]]
+<strong>Explanation:</strong> The answer [[-2,4],[3,3]] would also be accepted.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= k &lt;= points.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
+</ul>

1014-k-closest-points-to-origin/solution.py

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+# Approach 2: Max Heap
+
+# Time: O(n * log k)
+# Space: O(k)
+
+import heapq
+
+class Solution:
+    def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
+        heap = [(-self.squared_distance(points[i]), i) for i in range(k)]
+        heapq.heapify(heap)
+
+        for i in range(k, len(points)):
+            dist = -self.squared_distance(points[i])
+            if dist > heap[0][0]:
+                heapq.heappushpop(heap, (dist, i))
+
+        return [points[i] for (_, i) in heap]
+
+
+    def squared_distance(self, point):
+        return point[0] ** 2 + point[1] ** 2
+