Implement conflict detection

Sources/BijectiveDictionary/BijectiveDictionary+Conflict.swift

@@ -0,0 +1,86 @@
+//  =============================================================
+//  File: BijectiveDictionary+Conflict.swift
+//  Project: BijectiveDictionary
+//  -------------------------------------------------------------
+//  Created by Jacob Gelman on 09/11/2024
+//  Copyright © 2024 Jacob Gelman. All rights reserved.
+//  =============================================================
+
+extension BijectiveDictionary {
+
+    /// The result of a conflict check.
+    @frozen
+    public enum Conflict: Hashable {
+
+        /// The left value is already present.
+        case left
+
+        /// The right value is already present.
+        case right
+
+        /// Both the left and right values are already present in the same pair.
+        case pair
+
+        /// Both the left and right values are already present across two different pairs.
+        case both(otherLeft: Left, otherRight: Right)
+    }
+
+    /// Check if a conflict exists between the given pair and the contents of the dictionary that, if inserted,
+    /// would override an existing pair or break the bijective property.
+    ///
+    /// - Parameter pair: The left-right pair to perform the conflict check against.
+    /// - Returns: The conflict, if one exists, or `nil`, indicating neither the left nor right value already
+    ///   exist in the dictionary.
+    /// - Complexity: O(1)
+    ///
+    ///  The following example demonstrates creating a dictionary mapping element symbols
+    ///  to their atomic numbers and checking to see if a conflict exists:
+    ///  ```swift
+    ///  let dict: BijectiveDictionary = ["Ti": 22, "Si": 14, "He": 2]
+    ///
+    ///  guard let conflict = dict.conflict(for: ("Ti", 2)) else {
+    ///      print("No conflict")
+    ///      return
+    ///  }
+    ///  switch conflict {
+    ///  case .left:
+    ///      print("Symbol already present")
+    ///  case .right:
+    ///      print("Atomic number already present")
+    ///  case .pair:
+    ///      print("Pair already exists")
+    ///  case .both(let otherSymbol, let otherNumber):
+    ///      print("Other symbol: \(otherSymbol), other number: \(otherNumber)")
+    /// }
+    /// // prints "Other symbol: He, other number: 22"
+    /// ```
+    @inlinable
+    public func conflict(with pair: Element) -> Conflict? {
+        let existing = (findByLeft(pair.left), findByRight(pair.right))
+        return switch existing {
+        case (nil, nil): nil
+        case (nil, .some): .right
+        case (.some, nil): .left
+        case (.some(let byLeft), .some(let byRight)):
+            if byLeft.left != byRight.left || byLeft.right != byRight.right {
+                .both(otherLeft: byRight.left, otherRight: byLeft.right)
+            } else {
+                .pair
+            }
+        }
+    }
+
+    /// Find a pair in the dictionary by left value.
+    @inlinable
+    internal func findByLeft(_ leftValue: Left) -> Element? {
+        guard let rightValue = self[left: leftValue] else { return nil }
+        return (leftValue, rightValue)
+    }
+
+    /// Find a pair in the dictionary by right value.
+    @inlinable
+    internal func findByRight(_ rightValue: Right) -> Element? {
+        guard let leftValue = self[right: rightValue] else { return nil }
+        return (leftValue, rightValue)
+    }
+}

Tests/BijectiveDictionaryTests/BijectiveDictionaryTests.swift

@@ -273,4 +273,14 @@ func testAThousand() {
         #expect(bijectiveDict[right: rightV] == leftV)
     }
 }
+
+@Test
+func conflict() {
+    let dict: BijectiveDictionary = ["A": 1, "B": 2, "C": 3]
+    #expect(dict.conflict(with: ("D", 4)) == nil)
+    #expect(dict.conflict(with: ("A", 0)) == .left)
+    #expect(dict.conflict(with: ("E", 1)) == .right)
+    #expect(dict.conflict(with: ("A", 1)) == .pair)
+    #expect(dict.conflict(with: ("A", 3)) == .both(otherLeft: "C", otherRight: 1))
+}
 #endif