I have two arrays
let a1 = [obj1, obj2, obj3]
let a2 = [obj3, obj2, obj1]
Assume that the array elements are custom objects which are not sortable. I just want to check if the Arrays contain the same elements, in any order.
I tried this:
if a1==a2 { print("S1 is the same as S2") }
else { print("S1 is not the same as S2") }
but I get "S1 is not the same as S2" as output.
All I could think of are two solutions
- Sort and compare (doesn't work if elements are not sortable, say complex numbers)
- Subtract one array from the other
Is there any built-in function or operation to check if they are equal, without considering order?
I found How do I check in Swift if two arrays contain the same elements regardless of the order in which those elements appear in? which has the answer only for sortable Arrays.
If the items are unique and equatable (as in your example), convert to a Set and compare the Sets:
let a1 = [1, 2, 3]
let a2 = [3, 2, 1]
Set(a1) == Set(a2)
With some futzing, this can be made to work for an arbitrary class:
class Person : NSObject {
let name: String
init(name:String) {self.name = name}
override func isEqual(other: AnyObject?) -> Bool {
return other is Person && (other as! Person).name == self.name
}
override var hashValue : Int {
return self.name.hashValue
}
}
let a1 = [Person(name:"Matt"), Person(name:"Sam")]
let a2 = [Person(name:"Sam"), Person(name:"Matt")]
Set(a1) == Set(a2)
You can convert them into Set or NSSet instances which are unsorted by definition and compare those. Better yet, instead of using arrays at all, consider using sets in the first place.
let a1 = [1, 4, 5]
let a2 = [4, 5, 1]
let s1 = NSSet(array: a1)
let s2 = NSSet(array: a2)
print(s1 == s2) // Prints "true"
If objects may appear multiple times in your arrays, you need to use a NSCountedSet instead which also counts how often each object occurs in a set:
let a1 = [1, 1, 2]
let a2 = [1, 2, 2]
let s1 = NSSet(array: a1)
let s2 = NSSet(array: a2)
print(s1 == s2) // Prints "true"
let c1 = NSCountedSet(array: a1)
let c2 = NSCountedSet(array: a2)
print(c1 == c2) // Prints "false"
These 2 solutions also work if the arrays have duplicates.
Scenario 1: elements are sortable
let a1 = [1, 2, 3]
let a2 = [3, 2, 1]
let equals = a1.sorted() == a2.sorted()
Time Complexity: We need to sort both arrays so time complexity is O(n * log n) + O( m * log m)
Space complexity: A temporary copy of both arrays is created so the required space is O(n) + O (m).
Where n is the number of elements in a1 and m is the number of elements in a2.
Scenario 2: elements are NOT sortable
func equals<Element:Equatable>(listA:[Element], listB:[Element]) -> Bool {
guard listA.count == listB.count else { return false }
var listB = listB
for a in listA {
if let indexB = listB.indexOf(a) {
listB.removeAtIndex(indexB)
} else {
return false
}
}
return listB.isEmpty
}
Time Complexity: We are iterating the first array and each time we perform a linear search into the second array so O(n * m)
Space Complexity: We created a copy of the second array so O(m)
