/** 973. K Closest Points to Origin https://leetcode.com/problems/k-closest-points-to-origin/ Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane and an integer k, return the k closest points to the origin (0, 0). The distance between two points on the X-Y plane is the Euclidean distance (i.e., √(x1 - x2)2 + (y1 - y2)2). You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in). Example 1: Input: points = [[1,3],[-2,2]], k = 1 Output: [[-2,2]] Explanation: 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) < 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]]. Example 2: Input: points = [[3,3],[5,-1],[-2,4]], k = 2 Output: [[3,3],[-2,4]] Explanation: The answer [[-2,4],[3,3]] would also be accepted. Constraints: 1. 1 <= k <= points.length <= 104 2. -104 < xi, yi < 104 */ use std::collections::BinaryHeap; pub struct Solution {} impl Solution { /* Solution: Priority Queue, Time:O(nlog(k)), Space:O(k) */ pub fn k_closest(points: Vec<Vec<i32>>, k: i32) -> Vec<Vec<i32>> { //default is max heap let mut heap: BinaryHeap<(i32, Vec<i32>)> = BinaryHeap::new(); for point in points { let dist = Self::distance(&point); heap.push((dist, point));//O(1) if heap.len() > k as usize { heap.pop();//remove the top which is the maximum, O(log(n)) } } let mut result:Vec<Vec<i32>> = Vec::new(); while let Some((dist,array)) = heap.pop() { result.push(array) } result } fn distance(array: &Vec<i32>) -> i32 { array[0] * array[0] + array[1] * array[1] } }