# Leetcode - k-closest-points-to-origin

We have a list of `points`

on the plane. Find the `K`

closest points to the origin `(0, 0)`

.

(Here, the distance between two points on a plane is the Euclidean distance.)

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 = 1Output:[[-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]].

This is a quick select problem.

def k_closest(points, k) distances = [] points.each {|p| distances << count_distance(p)} distance = select_k(distances, 0, distances.length - 1, k) result = [] points.each do |p| result << p if count_distance(p) <= distance end return result end def select_k(distances, left, right, k) if left >= right return distances[left] end pivot = left pivot = partition(distances, left, right) if k - 1 == pivot return distances[pivot] end if k - 1 < pivot return select_k(distances, left, pivot - 1, k) end return select_k(distances, pivot+1, right, k) end def partition(distances, left, right) if left >= right return left end pivot = left l = left+1 r = right while true while l <= right && distances[pivot] > distances[l] l+=1 end while r >= left && distances[pivot] < distances[r] r-=1 end if l >= r distances[pivot], distances[r] = distances[r], distances[pivot] return r end distances[l], distances[r] = distances[r], distances[l] l += 1 r -= 1 end end def count_distance(point) return point[0]*point[0] + point[1]*point[1] end

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