![]() ![]() Note: It's an exercise for you to design an efficient algorithm for the top() operation. ![]() So total number of queue operations = 2n − 1 and time complexity = O(n). The algorithm dequeues n elements from Q1 and enqueues n − 1 element to Q2, where n is the size of the stack. So rather than moving elements from Q2 to Q1, we just need to swap the reference of Q1 with Q2. The critical question is: in the 3rd step of the pop operation, do we really need to move all the elements from Q2 to Q1? If we observe closely, the answer is no: the Q1 will be empty at that stage, and elements in Q2 are in the same order as earlier in Q1. Optimized implementation of the pop operation Finally, we return the value stored in the variable stackTop as the output of the pop operation.Then, we run a loop until queue Q2 is empty and move all the elements from queue Q2 to queue Q1.We also decrement the value of stackSize by 1 to reflect the new size of the stack. We dequeue the last remaining element in queue Q1 and store it in a variable called stackTop.In this loop, we dequeue each element from queue Q1 and enqueue it into queue Q2. Otherwise, we run a loop until there is only one element left in queue Q1. If queue Q1 is empty, we throw an error message "stack underflow" and return.To achieve this, we can follow the following steps. Implementation of the pop operationĪs previously mentioned, to perform the pop operation, our goal is to remove the last element inserted into queue Q1 using another queue, Q2. We are performing a constant number of operations i.e. After inserting the element, we increment the value of stackSize by 1 to track the current size of the stack. This means we insert the element x at the back of queue Q1, which will become the top element of the stack. To add a new element to the stack, we enqueue it to the first queue, Q1. ![]() We define methods for stack operations: void push(int x) and int pop().We declare a variable called stackSize to track the current size of the stack.We declare two queues, Q1 and Q2, to simulate stack operations.Suppose we define a class called StackUsingQueue to implement a stack using queues. Therefore, to perform the pop operation using a single queue, we need to use two queues. To do this, we can use an additional queue to store the removed elements. However, before removing the last element, we need to store and maintain the order of the removed elements for future use. This would mean removing all the elements in the queue except for the last element, which is the top element of the stack. One solution to this problem could be to continuously remove elements from the queue until we reach the element at the back. The key question here is: how do we remove the element at the back of the queue? But the queue is a first-in, first-out (FIFO) data structure, which means we can only remove elements from the front of the queue. However, if we want to perform a pop operation, we need to remove the last element inserted into the queue, which is at the back of the queue. In this case, we can perform a push operation by simply inserting the new element at the back of the queue. Suppose we use a single queue to implement a stack. This can be achieved by manipulating the order of the elements in the queue. Therefore, to simulate a stack using a queue, we need to keep track of the last element inserted and make sure that it is the first element to be deleted. In a stack, we insert and delete elements from one end only, but in a queue, we insert elements at the back and delete elements from the front. We can understand the basic idea for implementing a stack using a queue by considering the order of insertion and deletion in both data structures. Using two queues: O(n) pop operation and O(1) push operation Using a single queue: O(1) pop operation and O(n) push operation. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |