Passing hashset itself as parameter for reduce in swift - swift

Question

Given a hashset h = [1,2,3,4,5] for example, the purpose is to count the number of unique element i such that h.contains(i+1).
I can write down swift code using reduce() as
h.reduce(0,{h.contains($1+1) ? $0 + 1 : $0})
But what if h is an array containing duplicates instead of a hashset? I first need to convert it into hashset and then using the above expression:
Set(h).reduce(0,{Set(h).contains($1+1) ? $0 + 1 : $0})
But in this way we calculated Set(h).count + 1 times of Set(h) as pointed out by #carpsen90, is there any way to write the code like
Set(h).reduce(0,{self.contains($1+1) ? $0 + 1 : $0})
without using a temporary variable to store Set(h)?

Answer 1

Every time you call Set(h) a new set is calculated, so in your example Set(h).reduce(0,{Set(h).contains($1+1) ? $0 + 1 : $0}) , Set(h) will be calculated h.count + 1 times. Having a variable let set = Set(h) is the way to go here :
let set = Set(h)
let result = set.reduce(0) {set.contains($1+1) ? $0 + 1 : $0}
He is an alternative way of getting the desired result :
Let's create a dictionary that indicates wether a number has appeared in h:
var dict = [Int: Bool].init(minimumCapacity: h.count)
for x in h {
if dict[x] == nil {
dict[x] = true
}
}
And then, for each h element, check that its successor appears in the dictionary :
var count = 0
for entry in dict {
if dict[entry.key + 1] == true {
count += 1
}
}
And you could check the result :
print(count) //4

Answer 2

The problem here is that your array might contain duplicates and to filter the duplicates the easiest way is to convert it into the Set. And the correct way to do that is to save the set in a new variable hence it is unavoidable.
Though you can still use the reduce method without converting your array into a set like this:
var tempH = [Int]()
let c = h.reduce(0) { (result, item) in
if (!tempH.contains(item)) {
tempH.append(item)
return h.contains(item+1) ? (result + 1) : result
}
else {return result}
}
But, as you can notice in above code, we have to use a temporary array to track our duplicates. Hence an extra variable seems unavoidable here. Though no Sets are being used in above code.