Suppose that a system to process requests has several queues, one for each customer service representative that is currently working. Each request has at least the following attributes: public class Request implements Comparable{ String description; String timestamp; ) where its compare to method compares requests based on their timestamp. The Queue class implements the usual enqueue (Request. item), dequeue (), get. Size(), and isEmpty() methods. Suppose that you have access to three queues initially declared as: Queue 91 = new QueueO; Queue 92 = new Queue(); Queue q3 = new Queue(); which represent the requests to be processed by three customer service representatives.
Suppose that one of the customer service representatives is no longer available, so the requests stored in 91, 92, and q3 must be divided as evenly as possible into two queues (p1, p2) according to their time stamps. Example: The following three queues: 91->("r9483","15;03"), ("r2532","13:55"), ("r8431","10:40") 92-> ("r3452","16:03"), ("r5234","14:05"), ("r1432","11:50") 93-> ("r3837","14:35"), ("r8742","13:10") are now divided into the following two queues: p1 -> ("r9483","15;03"), ("r5234","14:05"), ("r8742","13:10"), ("r8431","10:40") P2 -> ("r3452","16;03"), ("r3837","14:35"), ("r2532","13:55"), ("r1432","11:50") The way that this works is to move elements from the original queues to the destination ones alternating destinations, i. e. from the three dequeued elements from 91, 92, and q3, enqueue the one with the earliest time into queue p1, then from the three dequeued elements from 91, 92, and q3, enqueue the one with the earliest time into queue p2, then from the three dequeued elements from 91, 92, and q3, enqueue the one with the earliest time into queue p1, then back to p2, and keep repeating until 1, 42, and q3 are empty. Write java code to do this. Make sure to declare your queues p1 and p2.