Is there a way to make this statement shorter?
let number = 1
if number == 0 || number == 1 {
print("ok")
}
like
let number = 1
if number == (0 || 1) {
print("ok")
}
This one doesn't work but I was wondering if there is a shorter way that I am not aware of.
EDIT:
I am aware of the switch case that might be useful when many case appear, but in my case I was more looking for something to use with ternary operators:
let oldNumber = 1
let newNumber = (oldNumber = 0 || 1) ? 4 : 13
Boolean equation expressions are evaluated always providing left (number) and right (0) side.
Not really shorter but you could write
let number = 1
if (0...1).contains(number) {
print("ok")
}
just for variety:
let c = { (number:Int, args:Int...) -> Bool in return args.contains(number) }
let number = 1;
if c(number, 0,1,9,4) {
print("ok")
}
only shorter if you are going to reuse the closure, has the advantage over the 0...1 range because you can specify a disjoint set of numbers. Can be used for inline if statements etc quite neatly
Little poor but nevertheless.
let number = 1
print((0...1).contains(number) ? "ok" : "bad")
Or:
let a = 1
let b = (0...1).contains(a) ? 4 : 5
When checking for several possible cases the switch statement might be more concise.
let number = 1
switch (number) {
case 0, 1:
print("ok")
default:
print("something else")
}
Related
This question already has answers here:
Refactored Solution In Swift
(2 answers)
Closed 6 years ago.
I'm attempting to solve HackerRank's Hash Table Ransom Note challenge. There are 19 test cases and I'm passing all but two of time due to timeout on larger data sets (10,000-30,000 entries).
I'm given:
1) an array of words contained in a magazine and
2) an array of words for a ransom note. My objective is to determine if the words in the magazine can be used to construct a ransom note.
I need to have enough unique elements in the magazineWords to satisfy the quantity needed by noteWords.
I'm using this code to make that determination...and it takes FOREVER...
for word in noteWordsSet {
// check if there are enough unique words in magazineWords to put in the note
if magazineWords.filter({$0==word}).count < noteWords.filter({$0==word}).count {
return "No"
}
}
What is a faster way to accomplish this task?
Below is my complete code for the challenge:
import Foundation
var magazineWords = // Array of 1 to 30,000 strings
var noteWords = // Array of 1 to 30,000 strings
enum RegexString: String {
// Letters a to z, A to Z, 1 to 5 characters long
case wordCanBeUsed = "([a-zA-Z]{1,5})"
}
func matches(for regexString: String, in text: String) -> [String] {
// Hat tip MartinR for this
do {
let regex = try NSRegularExpression(pattern: regexString)
let nsString = text as NSString
let results = regex.matches(in: text, range: NSRange(location: 0, length: nsString.length))
return results.map { nsString.substring(with: $0.range)}
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
func canCreateRansomNote(from magazineWords: [String], for noteWords: [String]) -> String {
// figure out what's unique
let magazineWordsSet = Set(magazineWords)
let noteWordsSet = Set(noteWords)
let intersectingValuesSet = magazineWordsSet.intersection(noteWordsSet)
// constraints specified in challenge
guard magazineWords.count >= 1, noteWords.count >= 1 else { return "No" }
guard magazineWords.count <= 30000, noteWords.count <= 30000 else { return "No" }
// make sure there are enough individual words to work with
guard magazineWordsSet.count >= noteWordsSet.count else { return "No" }
guard intersectingValuesSet.count == noteWordsSet.count else { return "No" }
// check if all the words can be used. assume the regex method works perfectly
guard noteWords.count == matches(for: RegexString.wordCanBeUsed.rawValue, in: noteWords.joined(separator: " ")).count else { return "No" }
// FIXME: this is a processor hog. I'm timing out when I get to this point
// need to make sure there are enough magazine words to write the note
// compare quantity of word in magazine with quantity of word in note
for word in noteWordsSet {
// check if there are enough unique words in magazineWords to put in the note
if magazineWords.filter({$0==word}).count < noteWords.filter({$0==word}).count {
return "No"
}
}
return "Yes"
}
print(canCreateRansomNote(from: magazineWords, for: noteWords))
I don't know how to read from the test case on the contest website or what frameworks you are allowed. If Foundation is allowed, you can use NSCountedSet
import Foundation
let fileContent = try! String(contentsOf: URL(fileURLWithPath: "/path/to/file.txt"))
let scanner = Scanner(string: fileContent)
var m = 0
var n = 0
scanner.scanInt(&m)
scanner.scanInt(&n)
var magazineWords = NSCountedSet(capacity: m)
var ransomWords = NSCountedSet(capacity: n)
for i in 0..<(m+n) {
var word: NSString? = nil
scanner.scanUpToCharacters(from: .whitespacesAndNewlines, into: &word)
if i < m {
magazineWords.add(word!)
} else {
ransomWords.add(word!)
}
}
var canCreate = true
for w in ransomWords {
if ransomWords.count(for: w) > magazineWords.count(for: w) {
canCreate = false
break
}
}
print(canCreate ? "Yes" : "No")
It works by going through the input file one word at a time, counting how many times that word appears in the magazine and in the ransom note. Then if any word appear more frequently in the ransom note than in the magazine, it fails the test immediately. Run the 30,000 words test case in less than 1 second on my iMac 2012.
So, need you help, I have loops, where I'd like to find first positive element and it will be text of label and exit from all loops, but every time I get last element:
for j in getArrayOfAllTimes[i].timeForGetDifference()
{
switch j - timeNow() {
case let x where x > 0:
nextTimeLabel.text = String(j - timeNow())
break
default:
break
}
}
How get first element > 0?
The first(where:) method of Array
Instead of explicitly breaking out of a loop when a first element that fulfills som predicate is found, you could simply make use of the Array method first(where:).
Since you haven't provided us with a minimal, complete and verifiable example (which you should) we'll construct such an example:
/* Example setup */
struct Foo: CustomStringConvertible {
private let bar: Int
init(_ bar: Int) { self.bar = bar }
func timeForGetDifference() -> Int {
return bar
}
var description: String {
return String(bar)
}
}
func timeNow() -> Int { return 10 }
let getArrayOfAllTimes = [Foo(6), Foo(2), Foo(9), Foo(4), Foo(11), Foo(3), Foo(13)]
// nextTimeLabel: some UILabel
For the example as per above, we could set the text property of the nextTimeLabel as follows, using first(where:) to find the first element fulfilling our predicate, given that it exists (otherwise; will return nil in which case we will not enter the optional binding block below).
if let firstNonNegativeFoo = getArrayOfAllTimes
.first(where: { $0.timeForGetDifference() - timeNow() > 0 }) {
nextTimeLabel.text = String(describing: firstNonNegativeFoo) // 11
}
As to why your own approach does not work as intended: a break statement within a case of a switch statement will simply end the execution of the switch statement (not the loop which is one level above the switch statement.
From the Language Reference - Statements:
Break Statement
A break statement ends program execution of a loop, an if statement,
or a switch statement.
In you case, you've added the break statements as the last statements of each case: here, particularly, the break has truly no effect (since the switch statement would break out anyway, after exiting the case which it entered).
for i in 1...3 {
switch i {
case is Int: print(i); break // redundant 'break'
case _: ()
}
} // 1 2 3
// ... the same
for i in 1...3 {
switch i {
case is Int: print(i)
case _: ()
}
} // 1 2 3
for j in getArrayOfAllTimes[i].timeForGetDifference() {
bool a = NO;
switch j - timeNow() {
case let x where x > 0:
a = YES
nextTimeLabel.text = String(j - timeNow())
break
default:
break
}
if a == YES {
break;
}
}
I've been studying for a coding exam by doing the HackerRank test cases, for the most part I've been doing well, but I get hung up on some easy cases and you all help me when I can't see the solution. I'm working on this problem:
https://www.hackerrank.com/challenges/ctci-ransom-note
A kidnapper wrote a ransom note but is worried it will be traced back to him. He found a magazine and wants to know if he can cut out whole words from it and use them to create an untraceable replica of his ransom note. The words in his note are case-sensitive and he must use whole words available in the magazine, meaning he cannot use substrings or concatenation to create the words he needs.
Given the words in the magazine and the words in the ransom note, print Yes if he can replicate his ransom note exactly using whole words from the magazine; otherwise, print No.
Input Format
The first line contains two space-separated integers describing the respective values of (the number of words in the magazine) and (the number of words in the ransom note).
The second line contains space-separated strings denoting the words present in the magazine.
The third line contains space-separated strings denoting the words present in the ransom note.
Each word consists of English alphabetic letters (i.e., to and to ).
The words in the note and magazine are case-sensitive.
Output Format
Print Yes if he can use the magazine to create an untraceable replica of his ransom note; otherwise, print No.
Sample Input
6 4
give me one grand today night
give one grand today
Sample Output
Yes
Explanation
All four words needed to write an untraceable replica of the ransom note are present in the magazine, so we print Yes as our answer.
And here is my solution:
import Foundation
func main() -> String {
let v = readLine()!.components(separatedBy: " ").map{Int($0)!}
var a = [String](); var b = [String]()
if v[0] < v[1] { return "No"}
for i in 0 ..< 2 {
if i == 0 {
a = (readLine()!).components(separatedBy: " ")
} else { b = (readLine()!).components(separatedBy: " ") }
}
// Get list of elements that intersect in each array
let filtered = Set(a).intersection(Set(b))
// Map set to set of Boolean where true means set a has enough words to satisfy set b's needs
let checkB = filtered.map{ word in reduceSet(b, word: word) <= reduceSet(a, word: word) }
// If mapped set does not contain false, answer is Yes, else No
return !checkB.contains(false) ? "Yes" : "No"
}
func reduceSet(_ a: [String], word: String) -> Int {
return (a.reduce(0){ $0 + ($1 == word ? 1 : 0)})
}
print(main())
I always time out on three of the 20 test-cases with this solution. So the solution seems to solve all the test cases, but not within their required time constraints. These are great practice, but it's so extremely frustrating when you get stuck like this.
I should note that I use Sets and the Set(a).intersection(Set(b)) because when I tried mapping an array of Strings, half the test-cases timed out.
Any cleaner, or more efficient solutions will be greatly appreciated! Thank you!
Thanks to #Alexander - I was able to solve this issue using NSCountedSet instead of my custom reduce method. It's much cleaner and more efficient. Here is the solution:
import Foundation
func main() -> String {
let v = readLine()!.components(separatedBy: " ").map{Int($0)!}
var a = [String](); var b = [String]()
if v[0] < v[1] { return "No"}
for i in 0 ..< 2 {
if i == 0 {
a = (readLine()!).components(separatedBy: " ")
} else { b = (readLine()!).components(separatedBy: " ") }
}
let countA = NSCountedSet(array: a)
let countB = NSCountedSet(array: b)
let intersect = Set(a).intersection(Set(b))
let check = intersect.map{ countB.count(for: $0) <= countA.count(for: $0) }
return !check.contains(false) ? "Yes" : "No"
}
print(main())
Many thanks!
I took the leisure of making some improvements on your code. I put comments to explain the changes:
import Foundation
func main() -> String {
// Give more meaningful variable names
let firstLine = readLine()!.components(separatedBy: " ").map{Int($0)!}
let (magazineWordCount, ransomNoteWordCount) = (firstLine[0], firstLine[1])
// a guard reads more like an assertion, stating the affirmative, as opposed to denying the negation.
// it also
guard magazineWordCount > ransomNoteWordCount else { return "No" }
// Don't use a for loop if it only does 2 iterations, which are themselves hardcoded in.
// Just write the statements in order.
let magazineWords = readLine()!.components(separatedBy: " ")
let ransomNoteWords = readLine()!.components(separatedBy: " ") //You don't need ( ) around readLine()!
let magazineWordCounts = NSCountedSet(array: magazineWords)
let ransomNoteWordCounts = NSCountedSet(array: ransomNoteWords)
// intersect is a verb. you're looking for the noun, "intersection"
// let intersection = Set(a).intersection(Set(b))
// let check = intersect.map{ countB.count(for: $0) <= countA.count(for: $0) }
// You don't actually care for the intersection of the two sets.
// You only need to worry about exactly the set of words that
// exists in the ransom note. Just check them directly.
let hasWordWithShortage = ransomNoteWordCounts.contains(where: { word in
magazineWordCounts.count(for: word) < ransomNoteWordCounts.count(for: word)
})
// Don't negate the condition of a conditional expression. Just flip the order of the last 2 operands.
return hasWordWithShortage ? "No" : "Yes"
}
print(main())
with the comments removed:
import Foundation
func main() -> String {
let firstLine = readLine()!.components(separatedBy: " ").map{Int($0)!}
let (magazineWordCount, ransomNoteWordCount) = (firstLine[0], firstLine[1])
guard magazineWordCount > ransomNoteWordCount else { return "No" }
let magazineWords = readLine()!.components(separatedBy: " ")
let ransomNoteWords = readLine()!.components(separatedBy: " ")
let magazineWordCounts = NSCountedSet(array: magazineWords)
let ransomNoteWordCounts = NSCountedSet(array: ransomNoteWords)
let hasWordWithShortage = ransomNoteWordCounts.contains{ word in
magazineWordCounts.count(for: word) < ransomNoteWordCounts.count(for: word)
}
return hasWordWithShortage ? "No" : "Yes"
}
print(main())
It's simpler, and much easier to follow. :)
So I am trying to loop through an array and increase a two counters based on the index in the array. The array has boolean values in it and the counters are for said boolean values. My implementation is error filled, so I am trying to figure out whether my logic is incorrect or my implementation is incorrect. So an extra pair of eyes would help
var numTrue = 0
var numFalse = 0
var boolArray = [true, true, false, true]
for index in boolArray.enumerate() {
if index == false{
numFalse++
} else {
numTrue++
}
}
print("Count true: \(numTrue)")
print("Count false: \(numFalse)")
Like Unheilig wrote enumarate is returning a sequenceType.
You can also remove the enumerate call, which in my opinion make the code in your case more readable like this:
var numTrue = 0
var numFalse = 0
var boolArray = [true, true, false, true]
for index in boolArray {
if index == false{
numFalse++
} else {
numTrue++
}
}
print("Count true: \(numTrue)")
print("Count false: \(numFalse)")
Edit 1: Quick-Help Doc
For the next time If you have a similar problem or error popping up you can also use the Quick-Help Documentation by holding alt and hovering over the method than a questionmark appears and you can click on the method. A window will open with a description of the method and sometimes an example like in the case of enumerate(). See the screenshot below:
Edit 2: Improved Solution
Swift provides methods on collections in your case an array to reduce the amount of code.
In your case you can use the method filter() which returns a new array by filtering out elements from the array on which it's called. The only argument is a closure (read more about closures here) that returns a boolean and it will execute this closure once for each element in the array.
Swift automatically provides shorthand argument names to inline closures which can be used to refer to the values of the closure's arguments by the names $0, $1, $2 and so on (from the documentation).
So in your case $0 stands for each element beginning at index 0.
count returns the number of elements in you array, so in your case by using filter it only returns 3 because true appears 3 times. trueCounter = 3
For the falseCounter you can easily subtract the result of the trueCounter from boolArray.count which is falseCounter = 4 - 3 -> 1
var boolArray = [true, true, false, true]
let trueCounter = boolArray.filter({$0 == true}).count
let falseCounter = boolArray.count - trueCounter
print("Count true: \(trueCounter)") // Count true: 3
print("Count false: \(falseCounter)") // Count false: 1
The code does not compile because enumerate is returning you a SequenceType in form (n, x).
Change your code to the following:
var numTrue = 0
var numFalse = 0
let boolArray = [true, true, false, true]
//We are interested in value, ignoring the index
for (_, value) in boolArray.enumerate()
{
if value == false
{
numFalse++
}
else
{
numTrue++
}
}
Output:
Count true: 3
Count false: 1
Try this:
var numTrue = 0
var numFalse = 0
let boolArray = [true, true, false, true]
for index in boolArray {
if index {
numTrue += 1
} else {
numFalse += 1
}
}
print("Count true: \(numTrue)")
print("Count false: \(numFalse)")
Just to add some more swifty spice, here's a different approach.
numTrue = boolArray.filter{ $0 }.count
numFalse = boolArray.count - numTrue
It does have different runtime characteristics though (i.e. creates a new array on the filter operation) - so I'd not recommend it unconditionally.
I have an array of numbers typed Int.
I want to loop through this array and determine if each number is odd or even.
How can I determine if a number is odd or even in Swift?
var myArray = [23, 54, 51, 98, 54, 23, 32];
for myInt: Int in myArray{
if myInt % 2 == 0 {
println("\(myInt) is even number")
} else {
println("\(myInt) is odd number")
}
}
Use the % Remainder Operator (aka the Modulo Operator) to check if a number is even:
if yourNumber % 2 == 0 {
// Even Number
} else {
// Odd Number
}
or, use remainder(dividingBy:) to make the same check:
if yourNumber.remainder(dividingBy: 2) == 0 {
// Even Number
} else {
// Odd Number
}
Swift 5 adds the function isMultiple(of:) to the BinaryInteger protocol.
let even = binaryInteger.isMultiple(of: 2)
let odd = !binaryInteger.isMultiple(of: 2)
This function can be used in place of % for odd/even checks.
This function was added via the Swift Evolution process:
Preliminary Discussion in Swift Forum
Swift Evolution Proposal
Accepted Swift Evolution Implementation
Notably, isEven and isOdd were proposed but not accepted in the same review:
Given the addition of isMultiple(of:), the Core Team feels that isEven and isOdd offer no substantial advantages over isMultiple(of: 2).
Therefore, the proposal is accepted with modifications. isMultiple(of:) is accepted but isEven and isOdd are rejected.
If desired, those methods can be added easily through extension:
extension BinaryInteger {
var isEven: Bool { isMultiple(of: 2) }
var isOdd: Bool { !isEven }
}
"Parity" is the name for the mathematical concept of Odd and Even:
https://en.wikipedia.org/wiki/Parity_(mathematics)
You can extend the Swift BinaryInteger protocol to include a parity enumeration value:
enum Parity {
case even, odd
init<T>(_ integer: T) where T : BinaryInteger {
self = integer.isMultiple(of: 2) ? .even : .odd
}
}
extension BinaryInteger {
var parity: Parity { Parity(self) }
}
which enables you to switch on an integer and elegantly handle the two cases:
switch 42.parity {
case .even:
print("Even Number")
case .odd:
print("Odd Number")
}
You can use filter method:
let numbers = [1,2,3,4,5,6,7,8,9,10]
let odd = numbers.filter { $0 % 2 == 1 }
let even = numbers.filter { $0 % 2 == 0 }