Leetcode: 973. K Closest Points to Origin
Naive solution
- Primary idea: Calcualte the distances of all points, sort the result and pick the first k items.
- Time Complexity: O(n*logn): due to sorting
- Space Complexity: O(n)
func kClosest(_ points: [[Int]], _ k: Int) -> [[Int]] {
var res = [([Int], Double)]()
// Callculate the distances to origin from all points
for point in points{
var dis = sqrt(pow(Double(point[0]), 2) + pow(Double(point[1]), 2))
res.append((point, dis))
}
// Sort the distances
res = res.sorted(by: {$0.1 < $1.1})
// Pickup only the first k items
var res1 = res[0..<k]
// Return the result
return res1.map{$0.0}
}
print(kClosest([[1,3],[-2,2]], 1)) // [[-2,2]]
print(kClosest([[3,3],[5,-1],[-2,4]], 2)) // [[3, 3], [-2, 4]]
print(kClosest([[3, 3], [5, -1], [-2, 4]], 2)) // [[3, 3], [-2, 4]]
Using Heap
The followign Heap implementation is taken from swift-algorithm-club repo. But you can also use Heap data structure from swift-collections repo
public struct Heap<T> {
/** The array that stores the heap's nodes. */
var nodes = [T]()
/**
* Determines how to compare two nodes in the heap.
* Use '>' for a max-heap or '<' for a min-heap,
* or provide a comparing method if the heap is made
* of custom elements, for example tuples.
*/
private var orderCriteria: (T, T) -> Bool
/**
* Creates an empty heap.
* The sort function determines whether this is a min-heap or max-heap.
* For comparable data types, > makes a max-heap, < makes a min-heap.
*/
public init(sort: @escaping (T, T) -> Bool) {
self.orderCriteria = sort
}
/**
* Creates a heap from an array. The order of the array does not matter;
* the elements are inserted into the heap in the order determined by the
* sort function. For comparable data types, '>' makes a max-heap,
* '<' makes a min-heap.
*/
public init(array: [T], sort: @escaping (T, T) -> Bool) {
self.orderCriteria = sort
configureHeap(from: array)
}
/**
* Configures the max-heap or min-heap from an array, in a bottom-up manner.
* Performance: This runs pretty much in O(n).
*/
private mutating func configureHeap(from array: [T]) {
nodes = array
for i in stride(from: (nodes.count/2-1), through: 0, by: -1) {
shiftDown(i)
}
}
public var isEmpty: Bool {
return nodes.isEmpty
}
public var count: Int {
return nodes.count
}
/**
* Returns the index of the parent of the element at index i.
* The element at index 0 is the root of the tree and has no parent.
*/
@inline(__always) internal func parentIndex(ofIndex i: Int) -> Int {
return (i - 1) / 2
}
/**
* Returns the index of the left child of the element at index i.
* Note that this index can be greater than the heap size, in which case
* there is no left child.
*/
@inline(__always) internal func leftChildIndex(ofIndex i: Int) -> Int {
return 2*i + 1
}
/**
* Returns the index of the right child of the element at index i.
* Note that this index can be greater than the heap size, in which case
* there is no right child.
*/
@inline(__always) internal func rightChildIndex(ofIndex i: Int) -> Int {
return 2*i + 2
}
/**
* Returns the maximum value in the heap (for a max-heap) or the minimum
* value (for a min-heap).
*/
public func peek() -> T? {
return nodes.first
}
/**
* Adds a new value to the heap. This reorders the heap so that the max-heap
* or min-heap property still holds. Performance: O(log n).
*/
public mutating func insert(_ value: T) {
nodes.append(value)
shiftUp(nodes.count - 1)
}
/**
* Adds a sequence of values to the heap. This reorders the heap so that
* the max-heap or min-heap property still holds. Performance: O(log n).
*/
public mutating func insert<S: Sequence>(_ sequence: S) where S.Iterator.Element == T {
for value in sequence {
insert(value)
}
}
/**
* Allows you to change an element. This reorders the heap so that
* the max-heap or min-heap property still holds.
*/
public mutating func replace(index i: Int, value: T) {
guard i < nodes.count else { return }
remove(at: i)
insert(value)
}
/**
* Removes the root node from the heap. For a max-heap, this is the maximum
* value; for a min-heap it is the minimum value. Performance: O(log n).
*/
@discardableResult public mutating func remove() -> T? {
guard !nodes.isEmpty else { return nil }
if nodes.count == 1 {
return nodes.removeLast()
} else {
// Use the last node to replace the first one, then fix the heap by
// shifting this new first node into its proper position.
let value = nodes[0]
nodes[0] = nodes.removeLast()
shiftDown(0)
return value
}
}
/**
* Removes an arbitrary node from the heap. Performance: O(log n).
* Note that you need to know the node's index.
*/
@discardableResult public mutating func remove(at index: Int) -> T? {
guard index < nodes.count else { return nil }
let size = nodes.count - 1
if index != size {
nodes.swapAt(index, size)
shiftDown(from: index, until: size)
shiftUp(index)
}
return nodes.removeLast()
}
/**
* Removes the minimum value from the heap and returns it.
* This is only valid for min-heaps.
*/
@discardableResult public mutating func popMin() -> T? {
guard !nodes.isEmpty else { return nil }
if nodes.count == 1 {
return nodes.removeLast()
} else {
let minValue = nodes[0]
nodes[0] = nodes.removeLast()
shiftDown(0)
return minValue
}
}
/**
* Takes a child node and looks at its parents; if a parent is not larger
* (max-heap) or not smaller (min-heap) than the child, we exchange them.
*/
internal mutating func shiftUp(_ index: Int) {
var childIndex = index
let child = nodes[childIndex]
var parentIndex = self.parentIndex(ofIndex: childIndex)
while childIndex > 0 && orderCriteria(child, nodes[parentIndex]) {
nodes[childIndex] = nodes[parentIndex]
childIndex = parentIndex
parentIndex = self.parentIndex(ofIndex: childIndex)
}
nodes[childIndex] = child
}
/**
* Looks at a parent node and makes sure it is still larger (max-heap) or
* smaller (min-heap) than its childeren.
*/
internal mutating func shiftDown(from index: Int, until endIndex: Int) {
let leftChildIndex = self.leftChildIndex(ofIndex: index)
let rightChildIndex = leftChildIndex + 1
// Figure out which comes first if we order them by the sort function:
// the parent, the left child, or the right child. If the parent comes
// first, we're done. If not, that element is out-of-place and we make
// it "float down" the tree until the heap property is restored.
var first = index
if leftChildIndex < endIndex && orderCriteria(nodes[leftChildIndex], nodes[first]) {
first = leftChildIndex
}
if rightChildIndex < endIndex && orderCriteria(nodes[rightChildIndex], nodes[first]) {
first = rightChildIndex
}
if first == index { return }
nodes.swapAt(index, first)
shiftDown(from: first, until: endIndex)
}
internal mutating func shiftDown(_ index: Int) {
shiftDown(from: index, until: nodes.count)
}
}
// MARK: - Searching
extension Heap where T: Equatable {
/** Get the index of a node in the heap. Performance: O(n). */
public func index(of node: T) -> Int? {
return nodes.firstIndex(where: { $0 == node })
}
/** Removes the first occurrence of a node from the heap. Performance: O(n). */
@discardableResult public mutating func remove(node: T) -> T? {
if let index = index(of: node) {
return remove(at: index)
}
return nil
}
}
- Primary idea: Use min heap to reduce time complexity from the naive solution
- Time Complexity: O(n*logk)
- Space Complexity: O(n)
struct Point: Comparable {
let x: Int
let y: Int
let dist: Int
static func <(lhs: Point, rhs: Point) -> Bool {
return lhs.dist < rhs.dist
}
}
func kClosest(_ points: [[Int]], _ k: Int) -> [[Int]] {
var heap = Heap<Point>{ $0 < $1 } // min heap
for point in points {
let distance = point[0] * point[0] + point[1] * point[1]
heap.insert(Point(x: point[0], y: point[1], dist: distance))
}
var result: [[Int]] = []
var kCopy = k
while kCopy > 0 {
let points = heap.popMin()!
result.append([points.x, points.y])
kCopy -= 1
}
return result
}
print(kClosest([[1,3],[-2,2]], 1)) // [[-2,2]]
print(kClosest([[3,3],[5,-1],[-2,4]], 2)) // [[3, 3], [-2, 4]]
print(kClosest([[3, 3], [5, -1], [-2, 4]], 2)) // [[3, 3], [-2, 4]]