973. K Closest Points to Origin

/**
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]
    }
}