How do I check in Swift if two arrays contain the same elements regardless of the order in which those elements appear in?

Question

Let's say there are two arrays...

var array1 = ["a", "b", "c"]
var array2 = ["b", "c", "a"]

I'd like the result of the comparison of these two arrays to be true, and the following...

var array1 = ["a", "b", "c"]
var array2 = ["b", "c", "a", "d"]

...to be false. How can I achieve that in Swift? I tried to convert both arrays to sets but for some reason Set() keeps removing some (usually duplicated) objects that the array contains.

Any help would be appreciated.

Answer 1

Swift 3, 4

extension Array where Element: Comparable {
    func containsSameElements(as other: [Element]) -> Bool {
        return self.count == other.count && self.sorted() == other.sorted()
    }
}

// usage
let a: [Int] = [1, 2, 3, 3, 3]
let b: [Int] = [1, 3, 3, 3, 2]
let c: [Int] = [1, 2, 2, 3, 3, 3]

print(a.containsSameElements(as: b)) // true
print(a.containsSameElements(as: c)) // false

---

Answer 2

Swift Compare arrays

input:

let array1 = ["a", "b", "c"]
let array2 = ["b", "c", "a", "c"]

Case 1: Duplicates are important, then use built-in Array.sort() function which uses Introsort under the hood with O(n log n) complexity

let array1Sorted = array1.sorted() //a, b, c
let array2Sorted = array2.sorted() //a, b, c, c

if (array1Sorted.count == array2Sorted.count && array1Sorted == array2Sorted) {
    //they are identical
}

Case 2: Duplicates are not important - use Set, with O(n) complexity. The Set implements Hashable so the task is to implement the hash function for element

let set1 = Set(array1) //b, c, a
let set2 = Set(array2) //b, c, a
    
if (set1.count == set2.count && set1 == set2) {
    //they are identical
}

[Swift Collections]

Answer 3

Swift 5.2 Solution

var array1 = ["a", "b", "c"]
var array2 = ["b", "c", "a"]

if array1.sorted() == array2.sorted() {
    print("array 1 & array 2 are same")
}

Answer 4

you can do something like this:

  array1.sortInPlace()
  array2.sortInPlace()

  print(array1,array2)

  if array1 == array2 {
    print("equal")
  } else {
  print("not equal") 
  }

and if don't want change origional array we can do

 let sorted1 = array1.sort()
 let sorted2 = array2.sort()

  if sorted1 == sorted2 {
    print("equal")
  }else {
    print("not equal")
  }

Answer 5

I know this question is old, and it also didn't want to determine if array1 was a subset of array2. However, This works in Swift 5.3 and Xcode 12.3:

var array1 = ["a", "b", "c"]
var array2 = ["b", "c", "a", "d"]

print("array1 == array2? \(Set(array1) == Set(array2))")
print("array1 subset to array2? \(Set(array1).isSubset(of: Set(array2)))")

Answer 6

Create function to compare them:

func containSameElements(var firstArray firstArray: [String], var secondArray: [String]) -> Bool {
    if firstArray.count != secondArray.count {
        return false
    } else {
        firstArray.sortInPlace()
        secondArray.sortInPlace()
        return firstArray == secondArray
    }
}

Then:

var array1 = ["a", "a", "b"]
var array2 = ["a", "b", "a"]

var array3 = ["a", "b", "c"]
var array4 = ["b", "c", "a", "d"]

print(containSameElements(firstArray: array1, secondArray: array2)) //true
print(containSameElements(firstArray: array3, secondArray: array4)) //false
print(array1) //["a", "a", "b"]
print(array2) //["a", "b", "a"]
print(array3) //["a", "b", "c"]
print(array4) //["b", "c", "a", "d"]

Answer 7

Here is a solution that does not require the element to be Comparable, but only Equatable. It is much less efficient than the sorting answers, so if your type can be made Comparable, use one of those.

extension Array where Element: Equatable {
    func equalContents(to other: [Element]) -> Bool {
        guard self.count == other.count else {return false}
        for e in self{
          guard self.filter{$0==e}.count == other.filter{$0==e}.count else {
            return false
          }
        }
        return true
    }
}

Answer 8

Solution for Swift 4.1/Xcode 9.4:

extension Array where Element: Equatable {
    func containSameElements(_ array: [Element]) -> Bool {
        var selfCopy = self
        var secondArrayCopy = array
        while let currentItem = selfCopy.popLast() {
            if let indexOfCurrentItem = secondArrayCopy.index(of: currentItem) {
                secondArrayCopy.remove(at: indexOfCurrentItem)
            } else {
                return false
            }
        }
        return secondArrayCopy.isEmpty
    }
}

The main advantage of this solution is that it uses less memory than other (it always creates just 2 temporary arrays). Also, it does not require for Element to be Comparable, just to be Equatable.

Answer 9

If elements of your arrays are conforming to Hashable you can try to use the bag (it's like a set with the registration of each item amount). Here I will use a simplified version of this data structure based on Dictionary. This extension helps to create bag from array of Hashable:

extension Array where Element: Hashable {
    var asBag: [Element: Int] {
        return reduce(into: [:]) {
            $0.updateValue(($0[$1] ?? 0) + 1, forKey: $1)
        }
    }
}

Now you need to generate 2 bags from initial arrays and compare them. I wrapped it in this extension:

extension Array where Element: Hashable {
    func containSameElements(_ array: [Element]) -> Bool {
        let selfAsBag = asBag
        let arrayAsBag = array.asBag
        return selfAsBag.count == arrayAsBag.count && selfAsBag.allSatisfy {
            arrayAsBag[$0.key] == $0.value
        }
    }
}

This solution was tested with Swift 4.2/Xcode 10. If your current Xcode version is prior to 10.0 you can find the function allSatisfy of ArraySlice in Xcode9to10Preparation. You can install this library with CocoaPods.

Answer 10

if I have

    array1 = ["x", "y", "z"]
    array2 = ["a", "x", "c"]

I can do

    array1.filter({array2.contains($0})

to return ["x"]

and likewise

    array1.filter({!array2.contains($0)})

to return ["y", "z"]