ASSIGNMENT:        ...

ASSIGNMENT:         

1. What is the time complexity for the following operations in queue:
a. Time complexity of enqueue in (circular) array-based queue.
b. Time complexity of dequeue in (circular) array-based queue.
c. Time complexity of enqueue in (noncircular) array-based queue.
d. Time complexity of dequeue in (noncircular) array-based queue.
e. Time complexity of enqueue in link-based queue.
f. Time complexity of dequeue in link-based queue.
g. What is the time complexity of size() in a link-based queue with a count variable?
h. What is the time complexity of size() in a link-based queue without a count variable? (See part 2 of this assignment.)

2. For the queue implementations, the code discussed in class uses a count variable to keep track of the number of elements in the queue. Rewrite the link-based implementation without a count variable.

3. Implement the following function: obtain user input as a String, then output whether the input is a Palindrome. Complete this task by storing the characters of the string in a queue, as well as in a stack. Remove characters from both and compare. If they always match, it’s a palindrome.

These solutions may offer step-by-step problem-solving explanations or good writing examples that include modern styles of formatting and construction of bibliographies out of text citations and references. Students may use these solutions for personal skill-building and practice. Unethical use is strictly forbidden.

package adtqueue;

class QueueNode{
    char value;
    QueueNode next;

    public QueueNode(char value){
       this.value = value;
       next = null;
    }

    public char getValue(){
       return value;
    }

    public QueueNode getNext() {
       return next;
    }

    public void setNext(QueueNode next) {
       this.next = next;
    }
}

class ADTQueue{

    private QueueNode front, rear;

    public ADTQueue(){
       front = null;
       rear = null;
    }

    public void enqueue(char value){
       QueueNode node = new QueueNode(value);
       if(rear == null){
            rear = node;
            if(front == null){
                front = node;
            }
       }else {
            rear.setNext(node);
            rear = node;
       }
    }

    public char dequeue(){
       if(!isEmpty()){
            QueueNode d = front;
            if(front == rear){
                front = null;
                rear = null;
            }else {
                front = d.getNext();
            }
            return d.getValue();
       }else {
            return ' ';
       }
    }...