Alternating Two-Instance Singleton Pattern Problem
Problem Implement the singleton pattern with a twist. First, instead of storing one instance, store two instances. And in every even call of getInstance(), return the first instance and in every odd call of getInstance(), return the second instance. Solution Method 1 - Using Singleton Pattern Here is the explanation: instances: An array holding the two singleton instances. callCount: A counter to keep track of how many times getInstance is called. Private Constructor: Ensures that the class cannot be instantiated directly. synchronized getInstance: Synchronisation is used to make the method thread-safe. It ensures that only one thread can execute this method at a time, avoiding the creation of multiple instances in a multi-threaded environment. Modulo Operator (%): Used to alternate between the two singleton instances based on the call count. Lazy Initialization: The instances are created only when they are first needed. Code Java public class TwoInstanceSingleton { private static final TwoInstanceSingleton[] instances = new TwoInstanceSingleton[2]; private static int callCount = 0; // Private constructor to prevent instantiation private TwoInstanceSingleton() {} public static synchronized TwoInstanceSingleton getInstance() { int index = callCount % 2; // Lazily initialize the instance if (instances[index] == null) { instances[index] = new TwoInstanceSingleton(); } callCount++; return instances[index]; } // Example usage public static void main(String[] args) { TwoInstanceSingleton obj1 = TwoInstanceSingleton.getInstance(); TwoInstanceSingleton obj2 = TwoInstanceSingleton.getInstance(); TwoInstanceSingleton obj3 = TwoInstanceSingleton.getInstance(); TwoInstanceSingleton obj4 = TwoInstanceSingleton.getInstance(); System.out.println(obj1 == obj2); // false System.out.println(obj1 == obj3); // true System.out.println(obj2 == obj4); // true System.out.println(obj1 == obj4); // false } } ...