I have a array that looks like this:
var myArray = ["1one", "1two", "1three", "1four", "1five", "1six"]
And to get random, I use this:
var originalNames = [String]()
func getRandomName() -> String {
if (names.count == 0) {
names = originalNames
}
let randomNumber = Int(arc4random_uniform(UInt32(names.count)))
return names.removeAtIndex(randomNumber)
}
And I use it like this:
self.randomLabel.text = getRandomName()
As you can see, the array contains six different strings. The code that I am currently using, will return add the strings inside the array at random, but I only want to return the first 3 strings randomly. How can I do this?
You can try using
var originalNames = [String]()
func getRandomName() -> String {
if (names.count == 0) {
names = originalNames
}
let randomNumber = Int(arc4random_uniform(3))
return names[randomNumber]
}
so let randomNumber = Int(arc4random_uniform(3)) this will return random Int value upto 3.
var myArray = ["1one", "1two", "1three", "1four", "1five", "1six"]
var result:[String] = []
while result.count < 3 {
let randomNumber = Int(arc4random_uniform(UInt32(myArray.count)))
result.append(myArray.removeAtIndex(randomNumber))
}
print(result) // "["1two", "1one", "1three"]\n"
if you don't want to modify the original array just make a copy of it
let myArray = ["1one", "1two", "1three", "1four", "1five", "1six"]
var inputNames = myArray
var result:[String] = []
while result.count < 3 {
result.append(inputNames.removeAtIndex(Int(arc4random_uniform(UInt32(inputNames.count)))))
}
print(result) // "["1six", "1two", "1one"]\n"
Related
I'm getting an Xcode error
Global function 'zip' requires that 'String.Element' (aka 'Character') conform to 'Sequence'
on array[i] = zip(array[i], arrayOfZeros)
func expandedForm(_ num: Int) -> String {
let lenght = String(num).count
var array = Array(String(num))
var i = 0
while array.count > i {
let numberOfZeros = array.count - 1 - i
let arrayOfZeros = Array(repeating: "0", count: numberOfZeros)
let string = array[i]
array[i] = zip(array[i], arrayOfZeros)
i += 1
}
return ""
}
I'm trying to merge two array of strings, but I think they are not the same type. Need some help on this.
func expandedForm(_ num: Int) -> String {
let lenght = String(num).count
var array = Array(String(num))
var i = 0
while array.count > i {
let numberOfZeros = array.count - 1 - i
let arrayOfZeros = Array(repeating: "0", count: numberOfZeros)
let string = array[i]
array[i] = zip(array[i], arrayOfZeros)
i += 1
}
return ""
}
You have some issues.
var array = Array(String(num)), if you pass your cursor on array, it's a [String.Element], ie a [Character], an array a Character.
So let string = array[i], that's misleading, because array[i] is a Character, not a String.
zip(_:_:) awaits for two sequences (~array), and you give as first parameter a Character.
Possible solutions:
Make array, a real array of String:
var array = Array(String(num)).map { String($0) }
And then:
array[i] = array[i] + arrayOfZeros.joined()
Or create a new variable var output that will be an array of String:
var output: [String] = []
And populate it:
output.append(String(array[i] + arrayOfZeros.joined())
I have the following array:
var notebookCovers = ["cover1", "cover2", "cover3", "cover4", "cover4", "cover6", "cover7", "cover8", "cover9", "cover10"]
and a UIButton that when it's pressed it generates a new UIImage with one of the elements of the array.
What I need to do is every time the button is tapped to generate random but unique element from the array (without repeating the elements) until they've all been selected and then restart the array again.
So far, I have it getting a random element but it's repeated and I cannot figure out how to it so it gets a unique image every time
func createNewNotebook() {
let newNotebook = Notebook()
let randomInt = randomNumber()
newNotebook.coverImageString = notebookCovers[randomInt]
notebooks.insert(newNotebook, at: 0)
collectionView.reloadData()
}
func randomNumber() -> Int {
var previousNumber = arc4random_uniform(UInt32(notebookCovers.count))
var randomNumber = arc4random_uniform(UInt32(notebookCovers.count - 1))
notebookCovers.shuffle()
if randomNumber == previousNumber {
randomNumber = UInt32(notebookCovers.count - 1)
}
previousNumber = randomNumber
return Int(randomNumber)
}
Set is a collection type that holds unique elements. Converting your notebooks array to Set also lets you take advantage of its randomElement function
var aSet = Set(notebooks)
let element = aSet.randomElement()
aSet.remove(element)
Copy the array. Shuffle the copy. Now just keep removing the first element until the copy is empty. When it is empty, start over.
Example:
let arr = [1,2,3,4,5]
var copy = [Int]()
for _ in 1...30 { // just to demonstrate what happens
if copy.isEmpty { copy = arr; copy.shuffle() }
let element = copy.removeFirst() ; print(element, terminator:" ")
}
// 4 2 3 5 1 1 5 3 2 4 4 1 2 3 5 1 4 5 2 3 3 5 4 2 1 3 2 4 5 1
If you want to create a looping solution:
let originalSet = Set(arrayLiteral: "a","b","c")
var selectableSet = originalSet
func repeatingRandomObject() -> String {
if selectableSet.isEmpty {
selectableSet = originalSet
}
return selectableSet.remove(selectableSet.randomElement()!)!
}
force unwrapping is kind of safe here, since we know that the result will never be nil. If you don't want to force unwrap:
let originalSet = Set(arrayLiteral: "a","b","c")
var selectableSet = originalSet
func repeatingRandomObject() -> String? {
if selectableSet.isEmpty {
selectableSet = originalSet
}
guard let randomElement = selectableSet.randomElement() else { return nil }
return selectableSet.remove(randomElement)
}
You can try something like this,
var notebookCovers = ["cover1", "cover2", "cover3", "cover4", "cover4", "cover6", "cover7", "cover8", "cover9", "cover10"]
var tappedNotebooks: [String] = []
func tapping() {
let notebook = notebookCovers[Int.random(in: 0...notebookCovers.count - 1)]
if tappedNotebooks.contains(notebook){
print("already exists trying again!")
tapping()
} else {
tappedNotebooks.append(notebook)
print("appended", notebook)
}
if tappedNotebooks == notebookCovers {
tappedNotebooks = []
print("cleared Tapped notebooks")
}
}
It is possible with shuffle:
struct AnantShuffler<Base: MutableCollection> {
private var collection: Base
private var index: Base.Index
public init?(collection: Base) {
guard !collection.isEmpty else {
return nil
}
self.collection = collection
self.index = collection.startIndex
}
public mutating func next() -> Base.Iterator.Element {
let result = collection[index]
collection.formIndex(after: &index)
if index == collection.endIndex {
collection.shuffle()
index = collection.startIndex
}
return result
}
}
fileprivate extension MutableCollection {
/// Shuffles the contents of this collection.
mutating func shuffle() {
let c = count
guard c > 1 else { return }
for (firstUnshuffled, unshuffledCount) in zip(indices, stride(from: c, to: 1, by: -1)) {
let d: Int = numericCast(arc4random_uniform(numericCast(unshuffledCount)))
let i = index(firstUnshuffled, offsetBy: d)
swapAt(firstUnshuffled, i)
}
}
}
Use:
let shuffler = AnantShuffler(collection: ["c1","c2","c3"].shuffled())
shuffler?.next()
I have an array as below with 13 items.
var data =[“MonthFolder/",
"MonthFolder/January",
"MonthFolder/February",
"MonthFolder/March",
"DayFolder/",
"DayFolder/Monday",
"DayFolder/Tuesday",
"DayFolder/Wednesday",
"DayFolder/Thursday",
"YearFolder/",
"YearFolder/2016",
"YearFolder/2015",
"YearFolder/2014"]
I am trying to print an array with end result as below
[("MonthFolder/January","MonthFolder/February","MonthFolder/March"),
("DayFolder/Monday","DayFolder/Tuesday","DayFolder/Wednesday","DayFolder/Thursday"),
("YearFolder/2016", "YearFolder/2015”, "YearFolder/2014")]”
I have done this
var fileName = [AnyObject]()
var allFiles = [AnyObject]()
for item in data{
let keyString = item.characters.last
if keyString != "/"{
fileName.append(item)
}
else if keyString == "/"{
if fileName.count > 0{
allFiles.append(fileName)
fileName = []
}
}
}
and got the output as
[("MonthFolder/January","MonthFolder/February","MonthFolder/March"),
("DayFolder/Monday","DayFolder/Tuesday","DayFolder/Wednesday","DayFolder/Thursday”)]
missing the third object in the allFiles because the else if condition fails for the last iteration. How do i achieve my desired output? Thank you.
By this way you will get expected output :
var fileName = [AnyObject]()
var allFiles = [AnyObject]()
for item in data{
let keyString = item.characters.last
if keyString != "/"{
fileName.append(item)
}
else if keyString == "/"{
if fileName.count > 0{
allFiles.append(fileName)
fileName = []
}
}
}
if fileName.count != 0 {
allFiles.append(fileName)
fileName = []
}
Hope this will help you :)
Try this
data.forEach { (item) -> () in
if item.characters.last != "/" {
fileName.append(item)
} else if fileName.count > 0 {
allFiles.append(fileName)
fileName = []
}
}
allFiles.append(fileName)
print(allFiles)
I have tried this
var data = ["MonthFolder/", "MonthFolder/January", "MonthFolder/February", "MonthFolder/March", "DayFolder/", "DayFolder/Monday", "DayFolder/Tuesday", "DayFolder/Wednesday", "DayFolder/Thursday", "YearFolder/", "YearFolder/2016", "YearFolder/2015", "YearFolder/2014"];
var fileName = []
for (item in data) {
var keyString = data[item].split("/");
if(keyString[1] == "")
{
var keyV = keyString[0];
fileName[keyV]= new Array();
}
else
{
fileName[keyV].push(data[item]);
}
}
console.log(fileName)
and i got output like
[
DayFolder
["DayFolder/Monday", "DayFolder/Tuesday", "DayFolder/Wednesday", "DayFolder/Thursday"]
MonthFolder
["MonthFolder/January", "MonthFolder/February", "MonthFolder/March"]
YearFolder
["YearFolder/2016", "YearFolder/2015", "YearFolder/2014"]
]
Please check
Functional way:
var data = ["MonthFolder/",
"MonthFolder/January",
"MonthFolder/February",
"MonthFolder/March",
"DayFolder/",
"DayFolder/Monday",
"DayFolder/Tuesday",
"DayFolder/Wednesday",
"DayFolder/Thursday",
"YearFolder/",
"YearFolder/2016",
"YearFolder/2015",
"YearFolder/2014"]
var allFolders = [String]()
var allFiles = [String]()
func gatherFiles(files: [String], inFolders folders: [String]) -> [[String]] {
return folders.map { folder in
return (files.filter { $0.containsString(folder)})
}
}
for item in data {
let components = item.componentsSeparatedByString("/").filter({!$0.isEmpty})
if components.count > 1 {
allFiles.append(item)
} else {
allFolders.append(item)
}
}
let finalArr = gatherFiles(allFiles, inFolders: allFolders)
Note
Cons: with this way, there's more iterations.
Pros: You could reuse it ?
You can also use more generic approach based on Dictionary struct
var filesDict = [String : [String]]()
for fileName in data {
let fileComponents = fileName.componentsSeparatedByString("/")
let fileKey = fileComponents[0]
let fileValue = fileComponents[1]
var itemsArray = filesDict[fileKey]
if itemsArray == nil {
itemsArray = []
}
if fileValue != "" {
itemsArray!.append(fileName)
}
filesDict[fileKey] = itemsArray
}
//conversion from Dictionary to Array
let outputArray = Array(filesDict.values)
print(outputArray)
output
[["MonthFolder/January", "MonthFolder/February", "MonthFolder/March"],
["DayFolder/Monday", "DayFolder/Tuesday", "DayFolder/Wednesday", "DayFolder/Thursday"],
["YearFolder/2016", "YearFolder/2015", "YearFolder/2014"]]
Can I put this in a loop instead of having to manually code it in for each chapter?
var chp1 = Array(Chapters[0].componentsSeparatedByString("\n"))
var chp2 = Array(Chapters[1].componentsSeparatedByString("\n"))
var chp3 = Array(Chapters[2].componentsSeparatedByString("\n"))
var chp4 = Array(Chapters[3].componentsSeparatedByString("\n"))
var chp5 = Array(Chapters[4].componentsSeparatedByString("\n"))
var chp6 = Array(Chapters[5].componentsSeparatedByString("\n"))
var chp7 = Array(Chapters[6].componentsSeparatedByString("\n"))
var chp8 = Array(Chapters[7].componentsSeparatedByString("\n"))
Yes you can use
var chp = []
for chapter in Chapters
{
chp.append(chapter.componentsSeparatedByString("\n")))
}
now you can access to the chapters like
chp[0]
chp[1]
chp[2]
Also you don't need to cast return value to Array.
for range in 0...8 {
var chp = Chapters[range]
let result = chp.componentsSeparatedByString("\n"))
}
Or
for (_, chapter) in Chapters.enumerate() {
let result = chapter.componentsSeparatedByString("\n"))
}
I am attempting to build an anagram checker for swift. This is my code. In case you don't know an anagram checker checks if two strings have the same characters in them but, order does not matter.
func checkForAnagram(#firstString: String, #secondString: String) -> Bool {
var firstStringArray: [Character] = []
var secondStringArray: [Character] = []
/* if case matters delete the next four lines
and make sure your variables are not constants */
var first = firstString
var second = secondString
first = first.lowercaseString
second = second.lowercaseString
for charactersOne in first {
firstStringArray += [charactersOne]
}
for charactersTwo in second {
secondStringArray += [charactersTwo]
}
if firstStringArray.count != secondStringArray.count {
return false
} else {
for elements in firstStringArray {
if secondStringArray.contains(elements){
return true
} else {
return false
}
}
}
}
var a = "Hello"
var b = "oellh"
var c = "World"
checkForAnagram(firstString: a, secondString: b)
I am getting an error message of.
'[Character]' does not have a member 'contains'
The accepted answer is compact and elegant, but very inefficient if compared to other solutions.
I'll now propose and discuss the implementation of a few variants of anagram checker. To measure performance, I'll use the different variants to find the anagrams of a given word out of an array of 50,000+ words.
// Variant 1: Sorting of Character
// Measured time: 30.46 s
func anagramCheck1(a: String, b: String) -> Bool {
return a.characters.sorted() == b.characters.sorted()
}
This is essentially the solution of the accepted answer, written in Swift 3 syntax. It's very slow because Swift's String, unlike NSString, is based on Character, which handles Unicode characters properly.
A more efficient solution exploits the NSCountedSet class, which allows us to represent a string as a set of characters, each with its own count. Two strings are anagrams if they map to the same NSCountedSet.
Note: checking string lengths as a precondition makes the implementation always more efficient.
// Variant 2: NSCountedSet of Character
// Measured time: 4.81 s
func anagramCheck2(a: String, b: String) -> Bool {
guard a.characters.count == b.characters.count else { return false }
let aSet = NSCountedSet()
let bSet = NSCountedSet()
for c in a.characters {
aSet.add(c)
}
for c in b.characters {
bSet.add(c)
}
return aSet == bSet
}
Better but not excellent. Here, one of the "culprits" is the use of the native Swift Character type (from Swift's String). Moving back to good old Objective-C types (NSString and unichar) makes things more efficient.
// Variant 3: NSCountedSet of unichar
// Measured time: 1.31 s
func anagramCheck3(a: String, b: String) -> Bool {
let aString = a as NSString
let bString = b as NSString
let length = aString.length
guard length == bString.length else { return false }
let aSet = NSCountedSet()
let bSet = NSCountedSet()
for i in 0..<length {
aSet.add(aString.character(at: i))
bSet.add(bString.character(at: i))
}
return aSet == bSet
}
Using NSCountedSet is fine, but before we compare two NSCountedSet objects, we fully populate them. A useful alternative is to fully populate the NSCountedSet for only one of the two strings, and then, while we populate the NSCountedSet for the other string, we fail early if the other string contains a character that is not found in the NSCountedSet of the first string.
// Variant 4: NSCountedSet of unichar and early exit
// Measured time: 1.07 s
func anagramCheck4(a: String, b: String) -> Bool {
let aString = a as NSString
let bString = b as NSString
let length = aString.length
guard length == bString.length else { return false }
let aSet = NSCountedSet()
let bSet = NSCountedSet()
for i in 0..<length {
aSet.add(aString.character(at: i))
}
for i in 0..<length {
let c = bString.character(at: i)
if bSet.count(for: c) >= aSet.count(for: c) {
return false
}
bSet.add(c)
}
return true
}
This is about the best timing we are going to get (with Swift). However, for completeness, let me discuss one more variant of this kind.
The next alternative exploits a Swift Dictionary of type [unichar: Int] to store the number of repetitions for each character instead of NSCountedSet. It's slightly slower than the previous two variants, but we can reuse it later to obtain a faster implementation.
// Variant 5: counting repetitions with [unichar:Int]
// Measured time: 1.36
func anagramCheck5(a: String, b: String) -> Bool {
let aString = a as NSString
let bString = b as NSString
let length = aString.length
guard length == bString.length else { return false }
var aDic = [unichar:Int]()
var bDic = [unichar:Int]()
for i in 0..<length {
let c = aString.character(at: i)
aDic[c] = (aDic[c] ?? 0) + 1
}
for i in 0..<length {
let c = bString.character(at: i)
let count = (bDic[c] ?? 0) + 1
if count > aDic[c] ?? 0 {
return false
}
bDic[c] = count
}
return true
}
Note that a vanilla Objective-C implementation using NSCountedSet, corresponding to Variant 3, is faster than all the previous versions by a rather large margin.
// Variant 6: Objective-C and NSCountedSet
// Measured time: 0.65 s
- (BOOL)anagramChecker:(NSString *)a with:(NSString *)b {
if (a.length != b.length) {
return NO;
}
NSCountedSet *aSet = [[NSCountedSet alloc] init];
NSCountedSet *bSet = [[NSCountedSet alloc] init];
for (int i = 0; i < a.length; i++) {
[aSet addObject:#([a characterAtIndex:i])];
[bSet addObject:#([b characterAtIndex:i])];
}
return [aSet isEqual:bSet];
}
Another way we can improve upon the previous attempts is to observe that, if we need to find the anagram of a given word, we might as well consider that word as fixed, and thus we could build the corresponding structure (NSCountedSet, Dictionary, ...) for that word only once.
// Finding all the anagrams of word in words
// Variant 7: counting repetitions with [unichar:Int]
// Measured time: 0.58 s
func anagrams(word: String, from words: [String]) -> [String] {
let anagrammedWord = word as NSString
let length = anagrammedWord.length
var aDic = [unichar:Int]()
for i in 0..<length {
let c = anagrammedWord.character(at: i)
aDic[c] = (aDic[c] ?? 0) + 1
}
let foundWords = words.filter {
let string = $0 as NSString
guard length == string.length else { return false }
var bDic = [unichar:Int]()
for i in 0..<length {
let c = string.character(at: i)
let count = (bDic[c] ?? 0) + 1
if count > aDic[c] ?? 0 {
return false
}
bDic[c] = count
}
return true
}
return foundWords
}
Now, in the previous variant we have counted with a [unichar:Int] Dictionary. This proves slightly more efficient than using an NSCountedSet of unichar, either with early exit (0.60 s) or without (0.87 s).
You should try
func checkForAnagram(firstString firstString: String, secondString: String) -> Bool {
return firstString.lowercaseString.characters.sort() == secondString.lowercaseString.characters.sort()
}
func checkAnagrams(str1: String, str2: String) -> Bool {
guard str1.count == str2.count else { return false }
var dictionary = Dictionary<Character, Int>()
for index in 0..<str1.count {
let value1 = str1[str1.index(str1.startIndex, offsetBy: index)]
let value2 = str2[str2.index(str2.startIndex, offsetBy: index)]
dictionary[value1] = (dictionary[value1] ?? 0) + 1
dictionary[value2] = (dictionary[value2] ?? 0) - 1
}
return !dictionary.contains(where: {(_, value) in
return value != 0
})
}
Time complexity - O(n)
// Make sure name your variables correctly so you won't confuse
// Mutate the constants parameter, lowercase to handle capital letters and the sorted them to compare both. Finally check is there are equal return true or false.
func anagram(str1: String, srt2: String)->Bool{
let string1 = str1.lowercased().sorted()
let string2 = srt2.lowercased().sorted()
if string1 == string2 {
return true
}
return false
}
// This answer also would work
// Convert your parameters on Array, then sorted them and compare them
func ana(str1: String, str2: String)->Bool{
let a = Array(str1)
let b = Array(str2)
if a.sorted() == b.sorted() {
return true
}
return false
}
Don't forget whitespaces
func isAnagram(_ stringOne: String, stringTwo: String) -> Bool {
return stringOne.lowercased().sorted().filter { $0 != " "} stringTwo.lowercased().sorted().filter { $0 != " "}
}
Swift 4.1 Function will give you 3 questions answer for Anagram :-
1. Input Strings (a,b) are Anagram ? //Bool
2. If not an Anagram then number of count require to change Characters in strings(a,b) to make them anagram ? // Int
3. If not an Anagram then list of Characters needs to be change in strings(a,b) to make them anagram ? // [Character]
STEP 1:- Copy and Paste below function in to your required class:-
//MARK:- Anagram checker
func anagramChecker(a:String,b:String) -> (Bool,Int,[Character]) {
var aCharacters = Array(a)
var bCharacters = Array(b)
var count = 0
var isAnagram = true
var replacementRequiredWords:[Character] = [Character]()
if aCharacters.count == bCharacters.count {
let listA = aCharacters.filter { !bCharacters.contains($0) }
for i in 0 ..< listA.count {
if !replacementRequiredWords.contains(listA[i]) {
count = count + 1
replacementRequiredWords.append(listA[i])
isAnagram = false
}
}
let listB = bCharacters.filter { !aCharacters.contains($0) }
for i in 0 ..< listB.count {
if !replacementRequiredWords.contains(listB[i]) {
count = count + 1
replacementRequiredWords.append(listB[i])
isAnagram = false
}
}
}else{
//cant be an anagram
count = -1
}
return (isAnagram,count,replacementRequiredWords)
}
STEP 2 :- Make two Input Strings for test
// Input Strings
var a = "aeb"
var b = "abs"
STEP 3:- Print results :-
print("isAnagram : \(isAnagram(a: a, b: b).0)")
print("number of count require to change strings in anagram : \(isAnagram(a: a, b: b).1)")//-1 will come in case of cant be a Anagram
print("list of Characters needs to be change : \(isAnagram(a: a, b: b).2)")
Results of above exercise:-
isAnagram : false
number of count require to change strings in anagram : 2
list of Characters needs to be change : ["e", "s"]
Hope this 10 minutes exercise will give some support to my Swift
family for solving Anagram related problems easily. :)
We can use dictionary to construct a new data structure container. Then compare the value by key/character of the string.
func anagram(str1: String, str2 : String) -> Bool {
var dict1 = [Character: Int]()
var dict2 = [Character:Int]()
for i in str1 {
if let count = dict1[i] {
dict1[i] = count + 1
} else {
dict1[i] = 1
}
}
for j in str2 {
if let count = dict2[j] {
dict2[j] = count + 1
} else {
dict2[j] = 1
}
}
return dict1 == dict2 ? true : false
}
// input -> "anna", "aann"
// The count will look like:
// ["a": 2, "n": 2] & ["a": 2, "n": 2]
// then return true
Another easy that I just realise doing an Anagram function in Swift 5.X
func checkForAnagram(firstString firstString: String, secondString: String) -> Bool {
return !firstString.isEmpty && firstString.sorted() == secondString.sorted()
}
class Solution {
func isAnagram(_ s: String, _ t: String) -> Bool {
guard s.count == t.count else { return false }
let dictS = s.reduce(into: [Character: Int]()) { $0[$1, default: 0] += 1 }
let dictT = t.reduce(into: [Character: Int]()) { $0[$1, default: 0] += 1 }
for letter in s {
if let count = dictS[letter] {
guard count == dictT[letter] else { return false }
}
}
return true
}
}
Check two strings are anagram using inout method in Swift
func checkAnagramString(str1: inout String, str2: inout String)-> Bool{
var result:Bool = false
str1 = str1.lowercased().trimmingCharacters(in: .whitespace)
str2 = str2.lowercased().trimmingCharacters(in: .whitespaces)
if (str1.count != str2.count) {
return result
}
for c in str1 {
if str2.contains(c){
result = true
}
else{
result = false
return result
}
}
return result
}
Call function to check strings are anagram or not
var str1 = "tommarvoloriddle"
var str2 = "iamlordvoldemort"
print(checkAnagramString(str1: &str1, str2: &str2)) //Output = true.
func isAnagram(word1: String, word2: String) -> Bool {
let set1 = Set(word1)
let set2 = Set(word2)
return set1 == set2
}
or
func isAnagram(word1: String,word2: String) -> Bool {
return word1.lowercased().sorted() == word2.lowercased().sorted()
}