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//===----------------------------------------------------------------------===//
//
// This source file is part of the Swift Collections open source project
//
// Copyright (c) 2022 - 2024 Apple Inc. and the Swift project authors
// Licensed under Apache License v2.0 with Runtime Library Exception
//
// See https://swift.org/LICENSE.txt for license information
//
//===----------------------------------------------------------------------===//
#if !COLLECTIONS_SINGLE_MODULE
import InternalCollectionsUtilities
#endif
extension TreeSet {
/// Returns a Boolean value that indicates whether this set is a subset of
/// the given set.
///
/// Set *A* is a subset of another set *B* if every member of *A* is also a
/// member of *B*, ignoring the order they appear in the two sets.
///
/// let a: TreeSet = [1, 2, 3, 4]
/// let b: TreeSet = [4, 2, 1]
/// b.isSubset(of: a) // true
///
/// - Parameter other: Another set.
///
/// - Returns: `true` if the set is a subset of `other`; otherwise, `false`.
///
/// - Complexity: Expected to be O(`self.count`) on average, if `Element`
/// implements high-quality hashing. The implementation is careful to make
/// the best use of hash tree structure to minimize work when possible,
/// e.g. by skipping over parts of the input trees.
@inlinable
public func isSubset(of other: Self) -> Bool {
self._root.isSubset(.top, of: other._root)
}
/// Returns a Boolean value that indicates whether this set is a subset of
/// the given set.
///
/// Set *A* is a subset of another set *B* if every member of *A* is also a
/// member of *B*, ignoring the order they appear in the two sets.
///
/// let a: TreeSet = [1, 2, 3, 4]
/// let b: TreeSet = [4, 2, 1]
/// b.isSubset(of: a) // true
///
/// - Parameter other: The keys view of a persistent dictionary.
///
/// - Returns: `true` if the set is a subset of `other`; otherwise, `false`.
///
/// - Complexity: Expected to be O(`self.count`) on average, if `Element`
/// implements high-quality hashing. The implementation is careful to make
/// the best use of hash tree structure to minimize work when possible,
/// e.g. by skipping over parts of the input trees.
@inlinable
public func isSubset<Value>(
of other: TreeDictionary<Element, Value>.Keys
) -> Bool {
self._root.isSubset(.top, of: other._base._root)
}
/// Returns a Boolean value that indicates whether this set is a subset of
/// the given sequence.
///
/// Set *A* is a subset of another set *B* if every member of *A* is also a
/// member of *B*, ignoring the order they appear in the two sets.
///
/// let a: TreeSet = [1, 2, 3, 4]
/// let b: TreeSet = [4, 2, 1]
/// b.isSubset(of: a) // true
///
/// - Parameter other: A sequence of elements, some of whose elements may
/// appear more than once.
///
/// - Returns: `true` if the set is a subset of `other`; otherwise, `false`.
///
/// - Complexity: In the worst case, this makes O(*n*) calls to
/// `self.contains` (where *n* is the number of elements in `other`),
/// and it constructs a temporary persistent set containing every
/// element of the sequence.
@inlinable
public func isSubset(of other: some Sequence<Element>) -> Bool {
if let other = _specialize(other, for: Self.self) {
return isSubset(of: other)
}
var it = self.makeIterator()
guard let first = it.next() else { return true }
if let match = other._customContainsEquatableElement(first) {
// Fast path: the sequence has fast containment checks.
guard match else { return false }
while let item = it.next() {
guard other.contains(item) else { return false }
}
return true
}
// FIXME: Would making this a BitSet of seen positions be better?
var seen: _Node = ._emptyNode()
for item in other {
let hash = _Hash(item)
guard _root.containsKey(.top, item, hash) else { continue }
guard seen.insert(.top, (item, ()), hash).inserted else { continue }
if seen.count == self.count {
return true
}
}
return false
}
}
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