1. Suppose that some application requires using two stacks whose elements are of the same type. A natural storage structure of such a two-stack data type would consist of two arrays and two top pointers. Explain why this may not be a space wise efficient implementation. A stack is a last in first out (LIFO) data type. A stack can be implemented using arrays where the data is stored in continuous memory locations. We have two operations in a stack push and pop. Since it has continuous memory allocated, if we try inserting data it gets saved in that free space present in either of the stacks. This is not a space wise efficient implementation because data is stored statically. In this case one stack can be full while the other is empty.

2. Using the basic queue and stack operations, write an algorithm to reverse the elements in a queue.
Create an empty stack
While the queue is not empty
Remove a value from the queue and push it onto the stack
While the stack is not empty
Pop a value from the stack and add it to the queue.

3. Assume that 'Stack' is the class described in this section with 'StackType' set to into and STACK_CAPACITY or myCapacity set to 5. Give the value of 'myTop' and the contents of the array referred to by 'myArray' in the Stack s afer the code segment is executed, or indicate why an error occurs.
Stack s;
s.push(1);
s.push(2);
s.push(3);
s.pop();
s.push(4);
s.push(5);
s.pop();
s.pop();

In a stack it is last in first out so
When we push items on to it we have 1 first then 2 and then 3
We have pop so, we pop 3 since it is the last element
Then again we have 4 then 5
Now we have two pops
Push 1,2,3 pop 3, push 4,5 pop 5,4 | | | | 5 | | 3 | 4 | | 2 | 2 | 2 | 1 | 1 | 1 |

Now we have 1 and 2 in the stack.