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1 |
| -# An implementation of the multiset data stucture |
| 1 | +# Multiset |
2 | 2 |
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3 |
| -## Usage |
| 3 | +A multiset (also known as a bag) is a data structure similar to a regular set, but it can store multiple instances of the same element. |
| 4 | + |
| 5 | +The following all represent the same *set*, but do not represent the same *multiset*. |
| 6 | + |
| 7 | +``` |
| 8 | +[1, 3, 2] |
| 9 | +[1, 3, 2, 2] |
| 10 | +``` |
| 11 | + |
| 12 | +In this multiset implementation, ordering is unimportant. So these two multisets are identical: |
| 13 | + |
| 14 | +``` |
| 15 | +[1, 2, 2] |
| 16 | +[2, 1, 2] |
| 17 | +``` |
| 18 | + |
| 19 | +Typical operations on a multiset are: |
| 20 | + |
| 21 | +- Add an element |
| 22 | +- Remove an element |
| 23 | +- Get the count for an element (the number of times it's been added) |
| 24 | +- Get the count for the whole set (the number of items that have been added) |
| 25 | +- Check whether it is a subset of another multiset |
| 26 | + |
| 27 | +## Implementation |
| 28 | + |
| 29 | +Under the hood, this implementation of Multiset uses a dictionary to store a mapping of elements to the number of times they've been added. |
| 30 | + |
| 31 | +Here's the essence of it: |
| 32 | + |
| 33 | +``` swift |
| 34 | +public struct Multiset<Element: Hashable> { |
| 35 | + fileprivate var storage: [Element: UInt] |
| 36 | + |
| 37 | + public init() { |
| 38 | + storage = [:] |
| 39 | + } |
| 40 | +``` |
| 41 | + |
| 42 | +And here's how you'd use this class to create a multiset of strings: |
| 43 | + |
| 44 | +``` swift |
| 45 | +var set = Multiset<String>() |
| 46 | +``` |
| 47 | + |
| 48 | +Adding an element is a case of incrementing the counter for that element, or setting it to 1 if it doesn't already exist: |
| 49 | + |
| 50 | +``` swift |
| 51 | +public mutating func add (_ elem: Element) { |
| 52 | + if let currentCount = storage[elem] { |
| 53 | + storage[elem] = currentCount + 1; |
| 54 | + } else { |
| 55 | + storage[elem] = 1 |
| 56 | + } |
| 57 | +} |
| 58 | +``` |
| 59 | + |
| 60 | +Here's how you'd use this method to add to the set we created earlier: |
| 61 | + |
| 62 | +```swift |
| 63 | +set.add("foo") |
| 64 | +set.add("foo") |
| 65 | +``` |
| 66 | + |
| 67 | +Our set now contains two elements, both the string "foo". |
| 68 | + |
| 69 | +Removing an element works much the same way as adding; decrement the counter for the element, or remove it from the underlying dictionary if its value is 1 before removal. |
| 70 | + |
| 71 | +``` swift |
| 72 | +public mutating func remove (_ elem: Element) { |
| 73 | + if let currentCount = storage[elem] { |
| 74 | + if currentCount > 1 { |
| 75 | + storage[elem] = currentCount - 1 |
| 76 | + } else { |
| 77 | + storage.removeValue(forKey: elem) |
| 78 | + } |
| 79 | + } |
| 80 | +} |
| 81 | +``` |
| 82 | + |
| 83 | +Getting the count for an item is simple: we just return the value for the given item in the internal dictionary. |
4 | 84 |
|
5 | 85 | ``` swift
|
6 |
| -var b = Multiset<Character>(); |
7 |
| -for c in "hello".characters { |
8 |
| - b.add(c) |
| 86 | +public func count(for key: Element) -> UInt { |
| 87 | + return storage[key] ?? 0 |
9 | 88 | }
|
10 |
| -let count = b.count // count is 5 |
11 |
| -let lcount = b.count(for: "l") // lcount is 2 |
12 | 89 | ```
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