1. Stack and Queue

2. Stack From Two Queues
Describe how to implement the stack ADT using two queues
The solution should only use the standard queue operations: ENQUEUE, DEQUEUE, IS-EMPTY
What will be the running times of the PUSH and POP operations?

3. Stack From Two Queues – Solution
We will call the two queues Q1 and Q2
Additionally, we will hold a variable called curQueue, which points to the queue that holds the last value pushed to our stack
Initialization: curQueue  Q1

4. Stack From Two Queues – Solution (continued)
IS-EMPTY
return IS-EMPTY (curQueue)
PUSH (val)
ENQUEUE (curQueue, val)
TOP
val  POP
return val

5. Stack From Two Queues – Solution (continued)
POP
otherQueue  (curQueue = Q1)? Q2: Q1
while (!IS-EMPTY (curQueue))
temp  DEQUEUE (curQueue)
if (!IS-EMPTY (curQueue)) // Not last?
ENQUEUE (otherQueue, temp)
curQueue  otherQueue
return temp

6. Stack From Two Queues – Solution Complexity
Since PUSH is simply implemented as a single ENQUEUE operation, its running time is O(1)
For POP operations, we always need to transfer the complete stack contents from one queue to the other, therefore performing a POP takes O(n) time

7. Two Stacks in a Single Array
Describe how to implement two stacks inside a single array
The total number of elements in both stacks is limited by the array length
Therefore, just splitting the array into two equally-sized parts will not do
All stack operations should run in O(1)

8. Two Stacks in a Single Array – Solution Concept
We are given an array A with n elements, numbered 0 to n – 1
We define two markers t1 and t2 that point to the next free cell in stacks S1 and S2 respectively
Initialization:
t1 = 0
t2 = n – 1

9. Two Stacks in a Single Array – Solution Concept
S1 grows up from the beginning of the array, S2 grows down from its end
We know that the array is full when the two markers switch places (t1 > t2)
The complexity is O(1) for all operations
The array usage is optimal

10. Two Stacks in a Single Array – Solution Visualization7 3 2 9 5 6 t1 t2

11. Two Stacks in a Single Array – Solution Details
S1.IS-EMPTY
if (t1 = 0)
return true
else
return false
S2.IS-EMPTY
if (t2 = n - 1)

12. Two Stacks in a Single Array – Solution Details (continued)
S1.PUSH (val)
if (t1 > t2)
error "stack overflow"
A[t1] = val
t1++
S2.PUSH (val)
A[t2] = val
t2--

13. Two Stacks in a Single Array – Solution Details (continued)
S1.POP
if (t1 = 0)
error "stack underflow"
t1--
return A[t1]
S2.POP
if (t2 = n - 1)
t2++
return A[t2]

14. Two Stacks in a Single Array – Solution Details (continued)
S1.TOP
if (t1 = 0)
error "stack is empty"
return A[t1]
S2.TOP
if (t2 = n - 1)
return A[t2]

15. Evaluating Arithmetic Expressions
Fully parenthesized arithmetic expressions can be evaluated using Dijkstra's two-stack algorithm
Uses two separate stacks – one for the operators and another for the operands
See Java code in: Evaluate.java.html

16. Deque (or Double-Ended Queue)
A deque is a generalization of the Queue ADT, that supports reading and writing from both ends of the queue
We would like to support an input-restricted deque :
Deletion can be made from both ends
Input can only be made at one end

17. Deque
We note that an input-restricted deque is a combination of a queue and a stack
We use a standard queue, and modify it to also support PUSH and POP
The tail (or rear) of the queue will also be used as the stack top

18. Deque Implementing PUSH is now trivial – simply call ENQUEUE
POP is very similar to DEQUEUE, accept that it returns tail instead of head
The same IS-EMPTY function will work for both the stack and the queue
Check if head = tail

19. Deque
ENQUEUE
PUSH
POP
DEQUEUE 9 5 6 2 3
head (front)
tail (rear)