Related
I want to find all the unique pairs whose sum will be equal to a given sum. I implement a solution but it returns only the first pair that it finds.
func checkPairs(in numbers: [Int], forSum target: Int) -> String {
for (i, x) in numbers.enumerated() {
for y in numbers[i+1 ..< numbers.count] {
if x + y == target {
return ("There is a pair that sums \(target): \(x)+\(y)=\(target)")
}
}
}
return "no pair"
}
print (checkPairs(in: [1, 2, 4, 4, 7, 5, 3], forSum: 8))
Output:
There is a pair that sums 8: 1+7=8
You exit the function early by calling return when the first pair is found. Instead, you should return an array of tuples containing the pairs, iterate over all pairs and add the ones satisfying the requirements to the return array.
func checkPairs(in numbers: [Int], forSum target: Int) -> [(Int,Int)] {
var pairs = [(Int,Int)]()
for (i, x) in numbers.enumerated() {
for y in numbers[i+1 ..< numbers.count] {
if x + y == target {
pairs.append((x,y))
}
}
}
return pairs
}
print(checkPairs(in: [1, 2, 4, 4, 7, 5, 3], forSum: 8))
Output:
[(1, 7), (4, 4), (5, 3)]
If you're looking for unique pairs, you can define a custom type holding the pair of numbers and return a Set instead of an Array (the custom type is needed, since you cannot make a Tuple conform to Hashable, which is needed to be able to add elements to a Set).
struct Pair: Hashable {
let smaller:Int
let bigger:Int
init(_ a:Int, _ b:Int) {
if a < b {
smaller = a
bigger = b
} else {
smaller = b
bigger = a
}
}
}
extension Pair: CustomStringConvertible {
var description:String {
return "\((smaller,bigger))"
}
}
func checkPairs(in numbers: [Int], forSum target: Int) -> Set<Pair> {
var pairs = Set<Pair>()
for (i, x) in numbers.enumerated() {
for y in numbers[i+1 ..< numbers.count] {
if x + y == target {
pairs.insert(Pair(x,y))
}
}
}
return pairs
}
print(checkPairs(in: [1, 2, 4, 4, 7, 5, 3], forSum: 8))
print(checkPairs(in: [1, 2, 4, 4, 7, 5, 3, 4, 1], forSum: 8))
Output:
[(4, 4), (3, 5), (1, 7)]
[(4, 4), (3, 5), (1, 7)]
here is one of options you can do
let array = [1, 2, 4, 4, 7, 5, 3]
let target = 8
func checkPairs(in numbers: [Int], forSum target: Int) -> String {
var pairs = ""
for (i, x) in numbers.enumerated() {
for y in numbers[i+1 ..< numbers.count] {
if x + y == target {
pairs += ("There is a pair that sums \(target): \(x)+\(y)=\(target)\n")
}
}
}
return pairs
}
print(checkPairs(in: array, forSum: target))
output will be There is a pair that sums 8: 1+7=8 There is a pair
that sums 8: 4+4=8 There is a pair that sums 8: 5+3=8
This question already has answers here:
Iterate over collection two at a time in Swift
(8 answers)
Closed 4 years ago.
I have an array of Int's. I need to change it in some way to get the new array of pairs of Int (array of tuples).
let a = [1, 2, 3, 4, 5]
// How to get array b which is
let b: [(Int, Int?)] // And it should be [(1, 2), (3, 4), (5, nil)]
The obvious solution is to use cycles. However I'm wondering whether it is more declarative way. Something like map which takes two arguments
// Something like this
let b = a.tupleMap { ($0, $1) }
UPDATE:
There is an awesome solution in related question.
Based on it I've added my own reduce method:
extension Collection {
func reduceByPairs<Result>(_ initialResult: Result, _ nextPartialResult: (Result, Element, Element?) throws -> Result) rethrows -> Result {
return try sequence(state: makeIterator()) { it in
it.next().map { ($0, it.next()) }
}
.reduce(initialResult) {
try nextPartialResult($0, $1.0, $1.1)
}
}
}
I would do something like this.
let array = [1, 2, 3, 4, 5]
let pairedArray = array
.enumerated()
.map { return ($0.element, array.count > $0.offset + 1 ? array[$0.offset + 1] : nil) }
.enumerated()
.filter({ $0.offset % 2 == 0 })
.map { $0.element }
print(pairedArray) // [(1, Optional(2)), (3, Optional(4)), (5, nil)]
Obviously the second element of the tuple is an Optional because you want to be able to have nil when there is no value.
You can try this:
extension Array {
func tupleMap(_ transform: (Element, Element?) -> (Element, Element?) = { ($0,$1) }) -> [(Element, Element?)] {
guard count > 0
else { return [] }
var result = [(Element, Element?)]()
var value1 = self[0]
var value2 : Element? = nil
for i in 0..<count {
if i % 2 == 0 {
value1 = self[i]
}
else {
value2 = self[i]
result.append(transform(value1, value2))
}
}
if count % 2 == 1 {
result.append(transform(value1, nil))
}
return result
}
}
Then you can use it this way:
let b = a.tupleMap { ($0, $1) }
Or even that way:
let b = a.tupleMap()
EDIT :
Or as proposed by Leo here: Iterate over collection two at a time in Swift
Here is a much shorter version:
extension Array {
func tupleMap(_ transform: (Element, Element?) -> (Element, Element?) = { ($0,$1) }) -> [(Element, Element?)] {
return stride(from: 0, to: count, by: 2).map {
transform(self[$0], $0 < count-1 ? self[$0.advanced(by: 1)] : nil)
}
}
}
Given array of n elements, i.e.
var array = [1, 2, 3, 4, 5]
I can write an extension to the Array so I can modify array to achieve this output: [2, 3, 4, 5, 1]:
mutating func shiftRight() {
append(removeFirst())
}
Is there a way to implement such a function that would shift array by any index, positive or negative. I can implement this function in imperative style with if-else clauses, but what I am looking for is functional implementation.
The algorithm is simple:
Split array into two by the index provided
append first array to the end of the second
Is there any way to implement it in functional style?
The code I've finished with:
extension Array {
mutating func shift(var amount: Int) {
guard -count...count ~= amount else { return }
if amount < 0 { amount += count }
self = Array(self[amount ..< count] + self[0 ..< amount])
}
}
You can use ranged subscripting and concatenate the results. This will give you what you're looking for, with names similar to the standard library:
extension Array {
func shiftRight(var amount: Int = 1) -> [Element] {
guard count > 0 else { return self }
assert(-count...count ~= amount, "Shift amount out of bounds")
if amount < 0 { amount += count } // this needs to be >= 0
return Array(self[amount ..< count] + self[0 ..< amount])
}
mutating func shiftRightInPlace(amount: Int = 1) {
self = shiftRight(amount)
}
}
Array(1...10).shiftRight()
// [2, 3, 4, 5, 6, 7, 8, 9, 10, 1]
Array(1...10).shiftRight(7)
// [8, 9, 10, 1, 2, 3, 4, 5, 6, 7]
Instead of subscripting, you could also return Array(suffix(count - amount) + prefix(amount)) from shiftRight().
With Swift 5, you can create shift(withDistance:) and shiftInPlace(withDistance:) methods in an Array extension with the following implementation in order to solve your problem:
extension Array {
/**
Returns a new array with the first elements up to specified distance being shifted to the end of the collection. If the distance is negative, returns a new array with the last elements up to the specified absolute distance being shifted to the beginning of the collection.
If the absolute distance exceeds the number of elements in the array, the elements are not shifted.
*/
func shift(withDistance distance: Int = 1) -> Array<Element> {
let offsetIndex = distance >= 0 ?
self.index(startIndex, offsetBy: distance, limitedBy: endIndex) :
self.index(endIndex, offsetBy: distance, limitedBy: startIndex)
guard let index = offsetIndex else { return self }
return Array(self[index ..< endIndex] + self[startIndex ..< index])
}
/**
Shifts the first elements up to specified distance to the end of the array. If the distance is negative, shifts the last elements up to the specified absolute distance to the beginning of the array.
If the absolute distance exceeds the number of elements in the array, the elements are not shifted.
*/
mutating func shiftInPlace(withDistance distance: Int = 1) {
self = shift(withDistance: distance)
}
}
Usage:
let array = Array(1...10)
let newArray = array.shift(withDistance: 3)
print(newArray) // prints: [4, 5, 6, 7, 8, 9, 10, 1, 2, 3]
var array = Array(1...10)
array.shiftInPlace(withDistance: -2)
print(array) // prints: [9, 10, 1, 2, 3, 4, 5, 6, 7, 8]
let array = Array(1...10)
let newArray = array.shift(withDistance: 30)
print(newArray) // prints: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
let array = Array(1...10)
let newArray = array.shift(withDistance: 0)
print(newArray) // prints: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
var array = Array(1...10)
array.shiftInPlace()
print(array) // prints: [2, 3, 4, 5, 6, 7, 8, 9, 10, 1]
var array = [Int]()
array.shiftInPlace(withDistance: -2)
print(array) // prints: []
I took a stab at writing some extensions for this. It has some nice features:
Shifting by an amount greater than count causes a wrap-around.
Shifting by negative amounts flips the direction
Exposes functions as the bit-shift binary operators (<<, <<=, >>, >>=)
extension Array {
public func shiftedLeft(by rawOffset: Int = 1) -> Array {
let clampedAmount = rawOffset % count
let offset = clampedAmount < 0 ? count + clampedAmount : clampedAmount
return Array(self[offset ..< count] + self[0 ..< offset])
}
public func shiftedRight(by rawOffset: Int = 1) -> Array {
return self.shiftedLeft(by: -rawOffset)
}
public mutating func shiftLeftInPlace(by rawOffset: Int = 1) {
if rawOffset == 0 { return /* no-op */ }
func shiftedIndex(for index: Int) -> Int {
let candidateIndex = (index + rawOffset) % self.count
if candidateIndex < 0 {
return candidateIndex + self.count
}
return candidateIndex
}
// Create a sequence of indexs of items that need to be swapped.
//
// For example, to shift ["A", "B", "C", "D", "E"] left by 1:
// Swapping 2 with 0: ["C", "B", "A", "D", "E"]
// Swapping 4 with 2: ["C", "B", "E", "D", "A"]
// Swapping 1 with 4: ["C", "A", "E", "D", "B"]
// Swapping 3 with 1: ["C", "D", "E", "A", "B"] <- Final Result
//
// The sequence here is [0, 2, 4, 1, 3].
// It's turned into [(2, 0), (4, 2), (1, 4), (3, 1)] by the zip/dropFirst trick below.
let indexes = sequence(first: 0, next: { index in
let nextIndex = shiftedIndex(for: index)
if nextIndex == 0 { return nil } // We've come full-circle
return nextIndex
})
print(self)
for (source, dest) in zip(indexes.dropFirst(), indexes) {
self.swapAt(source, dest)
print("Swapping \(source) with \(dest): \(self)")
}
print(Array<(Int, Int)>(zip(indexes.dropFirst(), indexes)))
}
public mutating func shiftRightInPlace(by rawOffset: Int = 1) {
self.shiftLeftInPlace(by: rawOffset)
}
}
public func << <T>(array: [T], offset: Int) -> [T] { return array.shiftedLeft(by: offset) }
public func >> <T>(array: [T], offset: Int) -> [T] { return array.shiftedRight(by: offset) }
public func <<= <T>(array: inout [T], offset: Int) { return array.shiftLeftInPlace(by: offset) }
public func >>= <T>(array: inout [T], offset: Int) { return array.shiftRightInPlace(by: offset) }
You can see it in action here.
Here is a more general solution, which implements this functionality lazily for any type that meets the requirements:
extension RandomAccessCollection where
Self: RangeReplaceableCollection,
Self.Index == Int,
Self.IndexDistance == Int {
func shiftedLeft(by rawOffset: Int = 1) -> RangeReplaceableSlice<Self> {
let clampedAmount = rawOffset % count
let offset = clampedAmount < 0 ? count + clampedAmount : clampedAmount
return self[offset ..< count] + self[0 ..< offset]
}
func shiftedRight(by rawOffset: Int = 1) -> RangeReplaceableSlice<Self> {
return self.shiftedLeft(by: -rawOffset)
}
mutating func shiftLeft(by rawOffset: Int = 1) {
self = Self.init(self.shiftedLeft(by: rawOffset))
}
mutating func shiftRight(by rawOffset: Int = 1) {
self = Self.init(self.shiftedRight(by: rawOffset))
}
//Swift 3
static func << (c: Self, offset: Int) -> RangeReplaceableSlice<Self> { return c.shiftedLeft(by: offset) }
static func >> (c: Self, offset: Int) -> RangeReplaceableSlice<Self> { return c.shiftedRight(by: offset) }
static func <<= (c: inout Self, offset: Int) { return c.shiftLeft(by: offset) }
static func >>= (c: inout Self, offset: Int) { return c.shiftRight(by: offset) }
}
Here's a functional implementation for "in place" rotation that doesn't require extra memory nor a temporary variable and performs no more than one swap per element.
extension Array
{
mutating func rotateLeft(by rotations:Int)
{
let _ = // silence warnings
(1..<Swift.max(1,count*((rotations+1)%(count+1)%1))) // will do zero or count - 1 swaps
.reduce((i:0,r:count+rotations%count)) // i: swap index r:effective offset
{ s,_ in let j = (s.i+s.r)%count // j: index of value for position i
swap(&self[j],&self[s.i]) // swap to place value at rotated index
return (j,s.r) // continue with next index to place
}
}
}
It optimally supports zero, positive and negative rotations as well as rotations of larger magnitude than the array size and rotation of an empty array (i.e. it cannot fail).
Uses negative values to rotate in the other direction (to the right).
Rotating a 3 element array by 10 is like rotating it by 1, the fist nine rotations will bring it back to its initial state (but we don't want to move elements more than once).
Rotating a 5 element array to the right by 3, i.e. rotateLeft(by:-3) is equivalent to rotateLeft(by:2). The function's "effective offset" takes that into account.
An easy solution,
public func solution(_ A : [Int], _ K : Int) -> [Int] {
if A.count > 0 {
let roundedK: Int = K % A.count
let rotatedArray = Array(A.dropFirst(A.count - roundedK) + A.dropLast(roundedK))
return rotatedArray
}
return []
}
I know I late to the party, but this answer based on the question works great?
extension Array {
mutating func shiftRight(p: Int) {
for _ in 0..<p {
append(removeFirst())
}
}
}
start [5, 0, 4, 11, 0]
shift [5, 0, 4, 11, 0] shift 0
shift [0, 4, 11, 0, 5] shift 1
shift [4, 11, 0, 5, 0] shift 2
shift [11, 0, 5, 0, 4] shift 3
Even better, if you ask it to shift more elements than there are in the array, it simply keeps circling.
Following the Nate Cook answers , I need also to shift an array returning reverse order, so I made:
//MARK: - Array extension
Array {
func shiftRight( amount: Int = 1) -> [Element] {
var amountMutable = amount
assert(-count...count ~= amountMutable, "Shift amount out of bounds")
if amountMutable < 0 { amountMutable += count } // this needs to be >= 0
return Array(self[amountMutable ..< count] + self[0 ..< amountMutable])
}
func reverseShift( amount: Int = 1) -> [Element] {
var amountMutable = amount
amountMutable = count-amountMutable-1
let a: [Element] = self.reverse()
return a.shiftRight(amountMutable)
}
mutating func shiftRightInPlace(amount: Int = 1) {
self = shiftRight(amount)
}
mutating func reverseShiftInPlace(amount: Int = 1) {
self = reverseShift(amount)
}
}
We have for example:
Array(1...10).shiftRight()
// [2, 3, 4, 5, 6, 7, 8, 9, 10, 1]
Array(1...10).shiftRight(7)
// [8, 9, 10, 1, 2, 3, 4, 5, 6, 7]
Array(1...10).reverseShift()
// [2, 1, 10, 9, 8, 7, 6, 5, 4, 3]
Array(1...10).reverseShift(7)
// [8, 7, 6, 5, 4, 3, 2, 1, 10, 9]
In objective C you can simply get left shifted array like this:
- (NSMutableArray *)shiftedArrayWithOffset:(NSInteger)offset
{
NSMutableArray *bufferArray = [[NSMutableArray alloc] initWithArray:originalArray];
for (int i = 0; i < offset; i++)
{
id object = [bufferArray firstObject];
[bufferArray removeObjectAtIndex:0];
[bufferArray addObject:object];
}
return bufferArray;
}
The fastest way is (but takes double memory!):
input:
var arr = [1,2,3,4,5]
let k = 1 (num steps to rotate)
let n = arr.count ( a little but faster )
rotation LEFT:
var temp = arr
for i in 0..<n {
arr[(n-i+k)%n] = temp[i]
}
result: [2, 1, 4, 3, 5]
rotation RIGHT:
var temp = arr
for i in 0..<n {
arr[(i+k)%n] = temp[i]
}
result: [4, 1, 2, 3, 5]
I came across a problem that required iterating over an array in pairs. What's the best way to do this? Or, as an alternative, what's the best way of transforming an Array into an Array of pairs (which could then be iterated normally)?
Here's the best I got. It requires output to be a var, and it's not really pretty. Is there a better way?
let input = [1, 2, 3, 4, 5, 6]
var output = [(Int, Int)]()
for i in stride(from: 0, to: input.count - 1, by: 2) {
output.append((input[i], input[i+1]))
}
print(output) // [(1, 2), (3, 4), (5, 6)]
// let desiredOutput = [(1, 2), (3, 4), (5, 6)]
// print(desiredOutput)
You can map the stride instead of iterating it, that
allows to get the result as a constant:
let input = [1, 2, 3, 4, 5, 6]
let output = stride(from: 0, to: input.count - 1, by: 2).map {
(input[$0], input[$0+1])
}
print(output) // [(1, 2), (3, 4), (5, 6)]
If you only need to iterate over the pairs and the given array is large
then it may be advantageous to avoid the creation of an intermediate
array with a lazy mapping:
for (left, right) in stride(from: 0, to: input.count - 1, by: 2)
.lazy
.map( { (input[$0], input[$0+1]) } ) {
print(left, right)
}
This is now available as
Sequence.chunks(ofCount: 2) of the swift-algorithms package
for chunk in input.chunks(ofCount: 2) {
print(chunk)
}
I don't think this is any better than Martin R's, but seems the OP needs something else...
struct PairIterator<C: IteratorProtocol>: IteratorProtocol {
private var baseIterator: C
init(_ iterator: C) {
baseIterator = iterator
}
mutating func next() -> (C.Element, C.Element)? {
if let left = baseIterator.next(), let right = baseIterator.next() {
return (left, right)
}
return nil
}
}
extension Sequence {
var pairs: AnySequence<(Self.Iterator.Element,Self.Iterator.Element)> {
return AnySequence({PairIterator(self.makeIterator())})
}
}
input.pairs.forEach{ print($0) }
let output = input.pairs.map{$0}
print(output) //->[(1, 2), (3, 4), (5, 6)]
Here's a version of #OOPer's answer that works with an odd number of elements in your list. You can leave off the conformance to CustomStringConvertible if you like, of course. But it gives prettier output for this example. : )
struct Pair<P: CustomStringConvertible>: CustomStringConvertible {
let left: P
let right: P?
var description: String {
if let right = right {
return "(\(left.description), \(right.description)"
}
return "(\(left.description), nil)"
}
}
struct PairIterator<C: IteratorProtocol>: IteratorProtocol where C.Element: CustomStringConvertible {
private var baseIterator: C
init(_ iterator: C) {
baseIterator = iterator
}
mutating func next() -> Pair<C.Element>? {
if let left = baseIterator.next() {
return Pair(left: left, right: baseIterator.next())
}
return nil
}
}
extension Sequence where Element: CustomStringConvertible {
var pairs: AnySequence<Pair<Self.Element>> {
return AnySequence({PairIterator(self.makeIterator())})
}
}
let input: [Int] = [1,2,3,4,5,6,7]
print(input.pairs)
print(Array(input.pairs))
//output:
AnySequence<Pair<Int>>(_box: Swift._SequenceBox<Swift._ClosureBasedSequence<__lldb_expr_27.PairIterator<Swift.IndexingIterator<Swift.Array<Swift.Int>>>>>)
[(1, 2, (3, 4, (5, 6, (7, nil)]
You don't need a custom type, like PairIterator as the above answers prescribe. Getting a paired sequence is a one-liner:
let xs = [1, 2, 3]
for pair in zip(xs, xs.dropFirst()) {
print(pair) // (1, 2) (2, 3)
}
If you intend to reuse that, you can place a pairs method inside an extension:
extension Sequence {
func pairs() -> AnySequence<(Element, Element)> {
AnySequence(zip(self, self.dropFirst()))
}
}
Given array of n elements, i.e.
var array = [1, 2, 3, 4, 5]
I can write an extension to the Array so I can modify array to achieve this output: [2, 3, 4, 5, 1]:
mutating func shiftRight() {
append(removeFirst())
}
Is there a way to implement such a function that would shift array by any index, positive or negative. I can implement this function in imperative style with if-else clauses, but what I am looking for is functional implementation.
The algorithm is simple:
Split array into two by the index provided
append first array to the end of the second
Is there any way to implement it in functional style?
The code I've finished with:
extension Array {
mutating func shift(var amount: Int) {
guard -count...count ~= amount else { return }
if amount < 0 { amount += count }
self = Array(self[amount ..< count] + self[0 ..< amount])
}
}
You can use ranged subscripting and concatenate the results. This will give you what you're looking for, with names similar to the standard library:
extension Array {
func shiftRight(var amount: Int = 1) -> [Element] {
guard count > 0 else { return self }
assert(-count...count ~= amount, "Shift amount out of bounds")
if amount < 0 { amount += count } // this needs to be >= 0
return Array(self[amount ..< count] + self[0 ..< amount])
}
mutating func shiftRightInPlace(amount: Int = 1) {
self = shiftRight(amount)
}
}
Array(1...10).shiftRight()
// [2, 3, 4, 5, 6, 7, 8, 9, 10, 1]
Array(1...10).shiftRight(7)
// [8, 9, 10, 1, 2, 3, 4, 5, 6, 7]
Instead of subscripting, you could also return Array(suffix(count - amount) + prefix(amount)) from shiftRight().
With Swift 5, you can create shift(withDistance:) and shiftInPlace(withDistance:) methods in an Array extension with the following implementation in order to solve your problem:
extension Array {
/**
Returns a new array with the first elements up to specified distance being shifted to the end of the collection. If the distance is negative, returns a new array with the last elements up to the specified absolute distance being shifted to the beginning of the collection.
If the absolute distance exceeds the number of elements in the array, the elements are not shifted.
*/
func shift(withDistance distance: Int = 1) -> Array<Element> {
let offsetIndex = distance >= 0 ?
self.index(startIndex, offsetBy: distance, limitedBy: endIndex) :
self.index(endIndex, offsetBy: distance, limitedBy: startIndex)
guard let index = offsetIndex else { return self }
return Array(self[index ..< endIndex] + self[startIndex ..< index])
}
/**
Shifts the first elements up to specified distance to the end of the array. If the distance is negative, shifts the last elements up to the specified absolute distance to the beginning of the array.
If the absolute distance exceeds the number of elements in the array, the elements are not shifted.
*/
mutating func shiftInPlace(withDistance distance: Int = 1) {
self = shift(withDistance: distance)
}
}
Usage:
let array = Array(1...10)
let newArray = array.shift(withDistance: 3)
print(newArray) // prints: [4, 5, 6, 7, 8, 9, 10, 1, 2, 3]
var array = Array(1...10)
array.shiftInPlace(withDistance: -2)
print(array) // prints: [9, 10, 1, 2, 3, 4, 5, 6, 7, 8]
let array = Array(1...10)
let newArray = array.shift(withDistance: 30)
print(newArray) // prints: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
let array = Array(1...10)
let newArray = array.shift(withDistance: 0)
print(newArray) // prints: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
var array = Array(1...10)
array.shiftInPlace()
print(array) // prints: [2, 3, 4, 5, 6, 7, 8, 9, 10, 1]
var array = [Int]()
array.shiftInPlace(withDistance: -2)
print(array) // prints: []
I took a stab at writing some extensions for this. It has some nice features:
Shifting by an amount greater than count causes a wrap-around.
Shifting by negative amounts flips the direction
Exposes functions as the bit-shift binary operators (<<, <<=, >>, >>=)
extension Array {
public func shiftedLeft(by rawOffset: Int = 1) -> Array {
let clampedAmount = rawOffset % count
let offset = clampedAmount < 0 ? count + clampedAmount : clampedAmount
return Array(self[offset ..< count] + self[0 ..< offset])
}
public func shiftedRight(by rawOffset: Int = 1) -> Array {
return self.shiftedLeft(by: -rawOffset)
}
public mutating func shiftLeftInPlace(by rawOffset: Int = 1) {
if rawOffset == 0 { return /* no-op */ }
func shiftedIndex(for index: Int) -> Int {
let candidateIndex = (index + rawOffset) % self.count
if candidateIndex < 0 {
return candidateIndex + self.count
}
return candidateIndex
}
// Create a sequence of indexs of items that need to be swapped.
//
// For example, to shift ["A", "B", "C", "D", "E"] left by 1:
// Swapping 2 with 0: ["C", "B", "A", "D", "E"]
// Swapping 4 with 2: ["C", "B", "E", "D", "A"]
// Swapping 1 with 4: ["C", "A", "E", "D", "B"]
// Swapping 3 with 1: ["C", "D", "E", "A", "B"] <- Final Result
//
// The sequence here is [0, 2, 4, 1, 3].
// It's turned into [(2, 0), (4, 2), (1, 4), (3, 1)] by the zip/dropFirst trick below.
let indexes = sequence(first: 0, next: { index in
let nextIndex = shiftedIndex(for: index)
if nextIndex == 0 { return nil } // We've come full-circle
return nextIndex
})
print(self)
for (source, dest) in zip(indexes.dropFirst(), indexes) {
self.swapAt(source, dest)
print("Swapping \(source) with \(dest): \(self)")
}
print(Array<(Int, Int)>(zip(indexes.dropFirst(), indexes)))
}
public mutating func shiftRightInPlace(by rawOffset: Int = 1) {
self.shiftLeftInPlace(by: rawOffset)
}
}
public func << <T>(array: [T], offset: Int) -> [T] { return array.shiftedLeft(by: offset) }
public func >> <T>(array: [T], offset: Int) -> [T] { return array.shiftedRight(by: offset) }
public func <<= <T>(array: inout [T], offset: Int) { return array.shiftLeftInPlace(by: offset) }
public func >>= <T>(array: inout [T], offset: Int) { return array.shiftRightInPlace(by: offset) }
You can see it in action here.
Here is a more general solution, which implements this functionality lazily for any type that meets the requirements:
extension RandomAccessCollection where
Self: RangeReplaceableCollection,
Self.Index == Int,
Self.IndexDistance == Int {
func shiftedLeft(by rawOffset: Int = 1) -> RangeReplaceableSlice<Self> {
let clampedAmount = rawOffset % count
let offset = clampedAmount < 0 ? count + clampedAmount : clampedAmount
return self[offset ..< count] + self[0 ..< offset]
}
func shiftedRight(by rawOffset: Int = 1) -> RangeReplaceableSlice<Self> {
return self.shiftedLeft(by: -rawOffset)
}
mutating func shiftLeft(by rawOffset: Int = 1) {
self = Self.init(self.shiftedLeft(by: rawOffset))
}
mutating func shiftRight(by rawOffset: Int = 1) {
self = Self.init(self.shiftedRight(by: rawOffset))
}
//Swift 3
static func << (c: Self, offset: Int) -> RangeReplaceableSlice<Self> { return c.shiftedLeft(by: offset) }
static func >> (c: Self, offset: Int) -> RangeReplaceableSlice<Self> { return c.shiftedRight(by: offset) }
static func <<= (c: inout Self, offset: Int) { return c.shiftLeft(by: offset) }
static func >>= (c: inout Self, offset: Int) { return c.shiftRight(by: offset) }
}
Here's a functional implementation for "in place" rotation that doesn't require extra memory nor a temporary variable and performs no more than one swap per element.
extension Array
{
mutating func rotateLeft(by rotations:Int)
{
let _ = // silence warnings
(1..<Swift.max(1,count*((rotations+1)%(count+1)%1))) // will do zero or count - 1 swaps
.reduce((i:0,r:count+rotations%count)) // i: swap index r:effective offset
{ s,_ in let j = (s.i+s.r)%count // j: index of value for position i
swap(&self[j],&self[s.i]) // swap to place value at rotated index
return (j,s.r) // continue with next index to place
}
}
}
It optimally supports zero, positive and negative rotations as well as rotations of larger magnitude than the array size and rotation of an empty array (i.e. it cannot fail).
Uses negative values to rotate in the other direction (to the right).
Rotating a 3 element array by 10 is like rotating it by 1, the fist nine rotations will bring it back to its initial state (but we don't want to move elements more than once).
Rotating a 5 element array to the right by 3, i.e. rotateLeft(by:-3) is equivalent to rotateLeft(by:2). The function's "effective offset" takes that into account.
An easy solution,
public func solution(_ A : [Int], _ K : Int) -> [Int] {
if A.count > 0 {
let roundedK: Int = K % A.count
let rotatedArray = Array(A.dropFirst(A.count - roundedK) + A.dropLast(roundedK))
return rotatedArray
}
return []
}
I know I late to the party, but this answer based on the question works great?
extension Array {
mutating func shiftRight(p: Int) {
for _ in 0..<p {
append(removeFirst())
}
}
}
start [5, 0, 4, 11, 0]
shift [5, 0, 4, 11, 0] shift 0
shift [0, 4, 11, 0, 5] shift 1
shift [4, 11, 0, 5, 0] shift 2
shift [11, 0, 5, 0, 4] shift 3
Even better, if you ask it to shift more elements than there are in the array, it simply keeps circling.
Following the Nate Cook answers , I need also to shift an array returning reverse order, so I made:
//MARK: - Array extension
Array {
func shiftRight( amount: Int = 1) -> [Element] {
var amountMutable = amount
assert(-count...count ~= amountMutable, "Shift amount out of bounds")
if amountMutable < 0 { amountMutable += count } // this needs to be >= 0
return Array(self[amountMutable ..< count] + self[0 ..< amountMutable])
}
func reverseShift( amount: Int = 1) -> [Element] {
var amountMutable = amount
amountMutable = count-amountMutable-1
let a: [Element] = self.reverse()
return a.shiftRight(amountMutable)
}
mutating func shiftRightInPlace(amount: Int = 1) {
self = shiftRight(amount)
}
mutating func reverseShiftInPlace(amount: Int = 1) {
self = reverseShift(amount)
}
}
We have for example:
Array(1...10).shiftRight()
// [2, 3, 4, 5, 6, 7, 8, 9, 10, 1]
Array(1...10).shiftRight(7)
// [8, 9, 10, 1, 2, 3, 4, 5, 6, 7]
Array(1...10).reverseShift()
// [2, 1, 10, 9, 8, 7, 6, 5, 4, 3]
Array(1...10).reverseShift(7)
// [8, 7, 6, 5, 4, 3, 2, 1, 10, 9]
In objective C you can simply get left shifted array like this:
- (NSMutableArray *)shiftedArrayWithOffset:(NSInteger)offset
{
NSMutableArray *bufferArray = [[NSMutableArray alloc] initWithArray:originalArray];
for (int i = 0; i < offset; i++)
{
id object = [bufferArray firstObject];
[bufferArray removeObjectAtIndex:0];
[bufferArray addObject:object];
}
return bufferArray;
}
The fastest way is (but takes double memory!):
input:
var arr = [1,2,3,4,5]
let k = 1 (num steps to rotate)
let n = arr.count ( a little but faster )
rotation LEFT:
var temp = arr
for i in 0..<n {
arr[(n-i+k)%n] = temp[i]
}
result: [2, 1, 4, 3, 5]
rotation RIGHT:
var temp = arr
for i in 0..<n {
arr[(i+k)%n] = temp[i]
}
result: [4, 1, 2, 3, 5]